Class XII Maths Chapter-06 - Application of Derivatives...Class XII – NCERT – Maths Chapter-06 - Application of Derivatives 6. Application of Derivatives Exercise 6.1. Question

Class XII – NCERT – Maths Chapter-06 - Application of Derivatives

6. Application of Derivatives

Exercise 6.1

Question 1:

Find the rate of change to the area of a circle with respect to its radius r when

(a) 3r cm (b) 4r cm

Solution 1:

The area of a circle A with radius r is given by,

2A r

Now, the area of the circle is changing of the area with respect to its radius is given by,

2 2dA d

r rdr dr

1. When 3r cm ,

2 3 6dA

dr

Hence, the area of the circle is changing at the rate of 6π cm when its radius is 3 .cm

2. When 4r cm ,

2 (4) 8dA

dr

Hence, the area of the circle is changing at the rate of 8π cm when its radius is 4 .cm

Question 2:

The volume of a cube is increasing at the rate of 38 cm /s. How fast is the surface area increasing

when the length of an edge is 12 cm?

Solution 2:

Let x be the length of a side, v be the volume, and s be the surface area of the cube.

Then, 3V x and 26S x when x is a function of time t .

It is given that 38 /dv

cm sdt

Then, by using the chain rule, we have:

3 28 . 3 .dv d dx dx

x x xdt x dt dt

2

8

3

dx

dt x … (1)

Now, 2 2(6 ) (6 )ds d d dx

x xdt dt dx dt

Bychain rule

2

8 32=12x. 12 .

3

dxx

dt x x

Thus, when 12 ,x cm 2 232 8/ / .

12 3

dScm s cm s

dt

https://www.ncertbooks.guru/cbse-ncert-solutions-pdf/



Hence, if the length of the edge of the cube is 12 ,cm then the surface area is increasing

at the rate of 8

3 cm2/s.

Question 3:

The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the

area of the circle is increasing when the radius is 10 cm.

Solution 3:


2A r

Now, the rate of change of area A with respect to time t is given by,

2 . 2dA d dr dr

r rdt d dx dt

By chain rule

It is given that,

3 /dr

cm sdt

2 3 6dA

r rdt

Thus, when 10 ,r cm

26 10 60 /dA

cm sdt

Hence, the rate at which the area of the circle is increasing when the radius is 10 cm is260 /cm s .

Question 4:

An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube

increasing when the edge is 10 cm long?

Solution 4:

Let x be the length of a side and v be the volume of the cube. Then, 3V x

23 .dV dx

xdt dt

by chain rule

It is given that,

3 /dx

cm sdt

2 23 3 9dV

x xdt

Then, when 10 ,x cm



2 39 10 900 /

dVcm s

dt

Hence, the volume of the cube is increasing at the rate of 900 cm3/s when the edge is 10 cm

long.

Question 5:

A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the

instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?

Solution 5:

The area of a circle (A) with radius (r) is given by 2A r

Therefore, the rate of change of area A with respect to time t is given by,

2 2 2dA d d dr dr

r r rdt dt dr dt dt

by chain rule

It is given that 5 /dr

cm sdt

Thus, when 8 ,r cm

2 8 5 80dA

dt

Hence, when the radius of the circular wave is 8 ,cm the enclosed area is increasing at the rate

of 280 / .cm s

Question 6:

The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its

circumference?

Solution 6:

The circumference of a circle C with radius r is given by

2 .C nr

Therefore, the rate of change of circumference C with respect to time t is given by,

.dC dC dr

dt dr dt by chain rule

2

2 .

d drr

dr dt

dr

dt

It is given that 0.7 /dr

cm sdt

Hence, the rate of increase of the circumference is

2 0.7 1.4 cm/s



Question 7:

The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing

at the rate of 4 cm/minute. When 8x cm and 6 ,y cm find the rates of change of a the

perimeter, and b the area of the rectangle.

Solution 7:

Since the length x is decreasing at the rate of 5 / minutecm and the width y is increasing

at the rate of 4 / minutecm , we have:

5 / min and

4 / min

dtcm

dy

dycm

dt

(a) The perimeter P of a rectangle is given by,

2P x y

2 2 5 4 2 cm/mindP dx dy

dt dt dt

Hence, the perimeter is decreasing at the rate of 2 cm/min,

(b) The area A of a rectangle is given by,

A x y

. . 5 4dA dx dy

y x y xdt dt dt

When 8x cm and 6 ,y cm 2 25 6 4 8 cm /min = 2 cm /mindA

dt

Hence, the area of the rectangle is increasing at the rate of 22cm /min .



Question 8:

A balloon, which always remains spherical on inflation, is being inflated by pumping in 900

cubic centimeters of gas per second. Find the rate at which the radius of the balloon increases

when the radius is 15 cm.

Solution 8:

The volume of a sphere V with radius r is given by,

34

3V r

Rate of change of volume V with respect to time t is given by,

.dV dV dr

dt dr dt by chain rule

3

2

4.

3

4 .

d drr

dr dt

drr

dt

It is given that

3900 /dv

cm Sdt

2900 4 .dr

rdt

2 2

900 225

4

dr

dt r r

Therefore, when radius = 15 cm,

2

225 1

15

dr

dt

Hence, the rate at which the radius of the balloon increases when the radius is 15 cm is 1

cm/s

Question 9:

A balloon, which always remains spherical has a variable radius. Find the rate at which its

volume is increasing with the radius when the later is 10 cm.

Solution 9:

The volume of a sphere v with radius r is given by 24

3V r

Rate of change of volume v with respect to its radius r is given by,

3 2 24 43 4

3 3

dV dr r r

dr dr

Therefore, when radius = 10 cm,

2

4 10 400dV

dr



Hence, the volume of the balloon is increasing at the rate of 400 3 / .cm s

Question 10:

A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground,

away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the

foot of the ladder is 4 m away from the wall?

Solution 10:

Let y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be

x m away from the wall.

Then, by Pythagoras Theorem, we have: 2 2 25x y Length of the ladder = 5 m

225y x

Then, the rate of change of height y with respect to time t is given by,

2.

25

dy x dx

dt dtx

It is given that 2 /dx

cm sdt

2

2

25

dy x

dt x

Now, when 4 ,x m we have:

2

2x4 8

325-4

dy

dt

Hence, the height of the ladder on the wall is decreasing at the rate of 8

3 cm/s.

Question 11:

A particle moving along the curve 36 2y x , Find the points on the curve at which the y

coordinate is changing 8 times as fast as the x-coordinate.

Solution 11:

The equation of the curve is given as: 36 2y x

The rate of change of the position of the particle with respect to time t , is given by,

2

2

6 3 0

2

dy dxx

dt dt

dy dxx

dt dt

When the y-coordinate of the particle changes 8 times as fast as the



x-coordinate i.e., 8dy dx

dt dt

, we have:

2

2

2

2

2 8

16

16 0

16

4

dx dxx

dt dt

dx dxx

dt dt

dxx

dt

x

x

When, 34 2 66

4, 116 6

x y

When,

3

4 2 62 314,

6 6 3x y

Hence, the points required on the curve are 4,11 and31

4,3

.

Question 12:

The radius of an air bubble is increasing at the rate of 1

2 cm/s. At what rate is the volume of the

bubbles of the bubble increasing when the radius is 1 cm?

Solution 12:

The air bubble is in the shape of a sphere.

Now, the volume of an air bubble V with radius r is given by, 24

3V r

The rate of change of volume V , with respect to time t is given by,

34.

3

dV d drr

dt dr dt By chain rule

2

2

43 .

3

4

drr

dt

drr

dt

It is given that 1

/2

drcm s

dt

Therefore, when 1 ,r cm

2 31

4 1 2 /2

dVcm s

dt

Hence, the rate at which the volume of the bubble increases in 32 / .cm s



Question 13:

A balloon, which always remains spherical, has a variable diameter 3

2 1 .2

x Find the rate of

change of its volume with respect to x .

Solution 13:

The volume of a sphere V with radius r is given by,

24

3V r

Diameter = 3

2 12

x

3(2 1)

4r x

3

3 34 4 3 9(2 1) (2 1)

3 3 4 16V r x x

Hence, the rate of change of volume with respect to x is as

3 2 39 9 27

2 1 x3 2 +1 x 2= 2 1 .16 16 8

dV dx x x

dt dt

Question 14:

Sand is pouring from a pipe at the rate of 312 /cm s . The falling sand forms a cone on the ground

in such a way that the height of the cone is always one-sixth of the radius of the base. How fast

is the height of the sand cone increasing when the height is 4 cm?

Solution 14:

The volume of a cone V with radius r and height h is given by.

21

3V r h

It is given that,

2 3

16

6

16 12

3

h r r h

V h h h

The rate of change of volume with respect to time t is given by,

312 .dV d dh

hdt dh dt

By chain rule

2

2

12 3

36

dhh

dt

dhh

dt

It is also given that 212 /dV

cm sdt



Therefore, when 4 ,h cm we have:

2

12 36 4dh

dt

12 1

36 16 48

dh

dt

Hence, when the height of the sand cone is 4 cm, its height is increasing at the rate of 1

48 cm/s.

Question 15:

The total cost C x in Rupees associated with the production of x units of an item is given by

3 20.007 0.003 15 4000C x x x x

Find the marginal cost when 17 units are produced.

Solution 15:

Marginal cost is the rate of change of total cost with respect to output.

Marginal cost 20.007 3 0.003 2 15dC

MC x xdx

20.021 0.006 15x x

When 17,x 20.021 17 0.006 17 15MC

0.021 289 0.006 17 15

6.069 0.102 15

20.967

Hence, when 17 units are produced, the marginal cost is Rs. 20.967.

Question 16:

The total revenue in Rupees received from the sale of x units of a product is given by

213 26 15R x x x

Find the marginal revenue when 7x .

Solution 16:

Marginal revenue is the rate of change of total revenue with respect to the number of units sold.

Marginal Revenue 13 2 26 26 26dR

MR x xdx

When 7,x

26 7 26 182 26 208MR

Hence, the required marginal revenue is Rs. 208.

Question 17:

The rate of change of the area of a circle with respect to its radius r at 6r cm is



(A) 10 (B) 12 (C) 8 (D) 11

Solution 17:


2A r

Therefore, the rate of change of the area with respect to its radius r is

2 2dA d

r rdt dr

When 6r cm ,

22 x6=12 cm /dA

sdr

Hence, the required rate of change of the area of a circle is 212 /cm s

The correct answer is B.

Question 18:

The total revenue in Rupees received from the sale of x units of a product is given by

23 36 5R x x x . The marginal revenue, when 15x is

(A) 116 (B) 96 (C) 90 (D) 126

Solution 18:

Marginal revenue is the rate of change of total revenue with respect to the number of units sold.

Marginal Revenue 3 2 36 6 36dR

MR x xdx

When 15,x

6 15 36 90 36 126MR

Hence, the required marginal revenue is Rs. 126

The correct answer is D.

Exercise 6.2

Question 1:

Show, that the function given by 3 17f x x is strictly increasing on .R

Solution 1:

Let 1x and 2x , be any two numbers in .R

1 2 1 2 1 2 2 23 3 3 17 3 17x x x x x x f x f x

Hence, f is strictly increasing on .R

Alternate Method:

f’(x) = 3 > 0, in every interval, on .R

Thus, the function is strictly increasing on .R



Question 2:

Show that the function given by 2xf x e is strictly increasing on .R

Solution 2:

Let 1x and 2x be any two numbers in .R

Then, we have:

1 22 2

1 2 1 2 1 22 2 x xx x x x e e f x f x

Hence, f is strictly increasing on .R

Question 3:

Show that the function given by sinf x x is

(A) Strictly increasing in 0,2

(B) Strictly decreasing ,2

(C) Neither increasing nor decreasing in ,O n

Solution 3:

The given function is sinf x x

' cosf x x

(A) Since for each 0,2

x

, cos 0,x we have ' 0f x .

Hence, f is strictly increasing in 0,2

.

(B) Since for each ,2

x

, cos 0,x we have ' 0f x .

Hence, f is strictly increasing in ,2

.

(C) From the results obtained in (A) and (B) it is clear that f is neither increasing nor decreasing

in 0,n .

Question 4:

Find the intervals in which the function f given by 22 3f x x x is

(A) Strictly increasing (B) strictly decreasing

Solution 4:

The given function is 22 3f x x x .



'

'

4 3

30

4

f x x

f x x

Now, the point 3

4 divides the real line into two disjoint intervals i.e.,

3,4

and 3

,4

.

In interval 3

,4

, ' 4 3 0.f x x

Hence, the given function f is strictly decreasing in interval3

,4

.

In interval 3

,4

, ' 4 3 0.f x x

Hence, the given function f is strictly increasing in interval3

,4

.

Question 5:

Find the intervals in which the function f given is

3 22 3 36 7f x x x x

(A) strictly increasing (B) strictly decreasing

Solution 5:

The given function is

3 22 3 36 7f x x x x

' 2 26 6 36 6 6 6 2 3f x x x x x x x

' 0f x 2,3x

The points 2x and 3x divide the real line into three disjoints intervals i.e.,

, 2 , 2,3 , and 3, .

In intervals , 2 and 3, , 'f x is positive while in interval

2,3 , 'f x is negative.

Hence, the given function f is strictly increasing in intervals

, 2 U 3, , while function f is strictly decreasing in interval 2,3 .



Question 6:

Find the intervals in which the following functions are strictly increasing or decreasing :

(a) 2 2 5x x

(b) 210 6 2x x

(c) 3 22 9 12 1x x x

(d) 26 9x x

(e) 3 3

1 3x x

Solution 6:

We have,

2 2 5f x x x

' 2 2f x x

Now,

' 0 1f x x

Point 1x divides the real line into two disjoint intervals i.e., , 1 and 1, .

In interval , 1 ,

' 2 2 0f x x

f is strictly decreasing in interval , 1 .

Thus, f is strictly decreasing for 1.x

In interval 1, ,

' 2 2 0f x x

f is strictly decreasing in interval 1, .

Thus f is strictly increasing for 1.x

(b) We have,

210 6 2f x x x

' 6 4f x x

Now,

' 30

2f x x

The point 3

2x divides the real line into two disjoint intervals

i.e.,3

,2

and 3

,2

.

In interval 3

,2

i.e., when3

2x , ' 6 4 0f x x .

f is strictly increasing for 3

2x .

In interval 3

,2

i.e., when3

2x , ' 6 4 0f x x .


2x .



(c) we have,

3 22 9 12 1f x x x x

' 2 26 18 12 6 3 2 6 1 2f x x x x x x x

Now,

' 0 1f x x and 2x

Points 1x and 2x divide the real line into three disjoint intervals.

i.e., , 2 , 2, 1 and 1,

in intervals , 2 and 1, i.e., when 2x and 1,x

' 6 1 2 0f x x x

f is strictly increasing for 2 1.x x

Now, in interval (–2, –1) i.e., when –2 < x < –1,

' 6 1 2 0.f x x x

f is strictly increasing for –2 < x < –1.

(d) We have,

26 9f x x x

' 9 2f x x

Now, 'f 0x gives9

2x

The point 9

2x divides the real line two disjoint intervals i.e.,

9,

2

and 9

,2

.

In interval 9

,2

i.e., for9

,2

x .


2x

In interval 9

,2

i.e., for

9

2x , ' 9 2 0.f x x

f is strictly decreasing for 9

2x

In the interval 9

,2

i.e., for '9, 9 2 0

2x f x x

f is strictly decreasing for x >9

2

(e) We have,

3 3

1 3f x x x

2 3 2 3

' 3 1 3 3 3 1f x x x x x x



2 2

2 2

2 2

3 1 3 3 1

3 1 3 2 2

6 1 3 1

x x x x

x x x

x x x

Now,

' 0 1,3,1f x x

The points 1x , 1x , and 3x divided the real line into four disjoint intervals.

i.e., , 1 , 1,1 , 1,3 and 3,

In intervals , 1 and 1,1 , 2 2' 6 1 3 1 0f x x x x

f is strictly decreasing in intervals , 1 and 1,1

In intervals 1,3 and 3, , 2 2' 6 1 3 1 0f x x x x

f is strictly increasing in intervals 1,3 and 3, .

Question 7:

Show that 2

log 1 , 12

xy x x

x

, is an increasing function of x throughout its domain.

Solution 7:

We have,

2

log 12

xy x

x

2

2 2 2

2 2 2 11 1 4

1 12 2 2

x xdy x

dx x xx x x

Now, 0dy

dx

2

20

2

x

x

2 0x 2 0 1x as x

0x

Since 1x , point 0x divides the domain 1, in two disjoint intervals i.e., 1 0x

and 0.x

When 1 0x we have:

2

2

0 0

1 2 0 2 0

x x

x x x

2'

20

2

xy

x

Also, when 0:x

220 0, 2 0x x x



2'

20

2

xy

x

Hence, function f is increasing throughout this domain.

Question 8:

Find the values of x for which 2

2y x x is an increasing function.

Solution 8:

We have,

2

2y x x = 2

2 2x x

22 2 2 2 4 2 1dy

x x x x x xdx

0 0, 2, 1dy

x x xdx

The points 0x , 1x and 2x divide the real line into four disjoint intervals i.e.,

,0 , 0,1 , 1,2 and 2, .

In intervals ,0 and 1,2 , 0dy

dx

y is strictly decreasing in intervals ,0 and 1,2

However, in intervals 0,1 and 2, , 0dy

dx

y is strictly decreasing in intervals 0,1 and 2,

y is strictly decreasing in intervals 0 1x and 2.x

Question 9:

Prove that

4sin

2 cosy

is an increasing function of in 0,

2

.

Solution 9:

We have,

4sin

2 cosy

2

(2 cos )(4cos ) 4sin ( sin )1

(2 cos )

dy

d

2 2

2

2

8cos 4cos 4sin1

2 cos

8cos 41

2 cos



Now,

0dy

d

2

8cos 41

2 cos

2

2

8cos 4 4 cos 4cos

cos 4cos 0

cos cos 4 0

cos 0 cos 4or

Since cos 4,cos 0.

cos 02

Now, 2 2

2 2

8cos (4 cos 4cos ) 4cos cos cos(4 cos )0

(2 cos ) 2 (2 cos ) (2 cos )

dy

d h

In interval 0,2

, we have cos 0 , Also 4 cos 4 cos 0 .

cos 4 cos 0 and also 2

2 cos 0

2

cos 4 cos0

2 cos

0dy

dx

Therefore, y is strictly increasing in interval 0,2

Also, the given function is continuous at 0x and 2

x

.

Hence, y is increasing in interval 0,2

.

Question 10:

Prove that the logarithmic function is strictly increasing on 0, .

Solution 10:

The given function is log .f x x

' 1f x

x

It is clear that for 0x , ' 10f x

x

Hence, logf x x is strictly increasing in interval 0, .



Question 11:

Prove that the function f is given by 2 1f x x x is neither strictly increasing nor strictly

decreasing on 1,1 .

Solution 11:

The given function is 2 1f x x x

' 2 1f x x

Now, ' 10 .

2f x x

The point 1

2 divides the interval 1,1 into two disjoint intervals i.e.,

11,

2

and 1

,12

.

Now, in interval 1

1,2

, ' 2 1 0.f x x

Therefore, f is strictly decreasing in interval 1

1,2

However, in interval 1

,12

, ' 2 1 0.f x x

Therefore, f is strictly increasing in interval 1

,12

.

Hence, f is neither strictly increasing nor decreasing in interval 1,1 .

Question 12:

Which of the following functions are strictly decreasing on 0, ?2

(A) cos x (B) cos 2x (C) cos 3x (D) tan x

Solution 12:

(A) let 1 cos .f x x

'

1 sinf x x

In interval 0,2

, '

1 sin 0.f x x

1 cosf x x is strictly decreasing in interval 0,2

.

(B) let 2 cos2f x x

'

2 2sin 2f x x

Now, 0 0 2 sin 2 0 2sin 2 02

x x x x



'

2 2sin 2 0 0,2

f x x on

2 cos2f x x is strictly decreasing in interval 0,2

.

(C) let 3 cos3f x x

'

3 3sin3f x x

Now, '3 0.f x

sin3 0 3 , 0,2

x x as x

3x

The point 3

x

divides the interval 0,2

into two disjoint intervals

i.e., 0,3

and , .3 2

Now, in interval 0,3

, 3 3sin3 0 0 0 33

f x x as x x

3f is strictly decreasing in interval 0,3

.

However, in interval ,3 2

, 3

33sin3 0 3

3 2 2f x x as x x

3f is strictly increasing in interval ,3 2

.

Hence, 3f is neither increasing nor decreasing in interval 0,2

.

(D) let 4 tanf x x

' 2

4 secf x x

In interval 0,2

' 2

4, sec 0.f x x

4f is strictly increasing in interval 0,2

.

Therefore, function cos x and cos2x are strictly decreasing in 0,2

Hence, the correct answer are A and B.

Question 13:

On which of the following intervals is the function f is given by 100 sin 1f x x x strictly

decreasing?



A. 0,1

B. ,2

C. 0,2

D. None of these

Solution 13:

We have,

100 sin 1f x x x

' 90100 cosf x x x

In interval, 0,1 , cos 0x and 90100 0x

' 0.f x

Thus, function f is strictly increasing in interval 0,1 .

In interval ,2

, 90cos 0 100 0x and x .Also 90100 cosx x

' 0 ,2

f x in

Thus, function f is strictly increasing in interval ,2

.

In interval 0,2

, 90cos 0 100 0x and x .

90100 cos 0x x

' 0 0,2

f x on

f is strictly increasing in interval 0,2

.

Hence, function f is strictly decreasing in none of the intervals.


Question 14:

Find the least value of a such that the function f given 2 1f x x ax is strictly increasing

on 1,2 .

Solution 14:

We have,

2 1f x x ax

' 2f x x a



Now, function f will be increasing in 1,2 if ' 0f x in 1,2 .

2 0

2

2

x a

x a

ax

Therefore, we have to find the least value of a such that

,2

ax

when 1,2x .

2

ax

(when 1 2x )

Thus, the least value of a for f to be increasing on 1,2 is given by,

12

a

1 22

aa

Hence, the required value of a is –2.

Question 15:

Let I be any interval disjoint from 1,1 , prove that the function f given by 1

f x xx

is

strictly increasing on I .

Solution 15:

We have,

1

f x xx

'

2

11f x

x

Now,

0 1f x x

The points 1x and 1x divided the real line in three disjoint intervals i.e.,

, 1 , 1,1 and 1, .

In interval 1,1 , it is observed that:

2

2

2

'

2

1 1

1

11 , 0

11 0, 0

11 0 1,1 ~ 0 .

x

x

xx

xx

f x onx



f is strictly decreasing on 1,1 ~ 0 .

In interval , 1 and 1, , it is observed that:

2

2

2

1 1

1

11

11 0

x or x

x

x

x

'

2

11 0f x

x on , 1 and 1, .

f is strictly increasing on , 1 and 1, .

Hence, function f is strictly increasing in interval in interval I disjoint from 1,1 .

Hence, the given result is proved.

Question 16:

Prove that the function f given by logsinf x x is strictly increasing on 0,2

and strictly

decreasing on ,2

Solution 16:

We have,

logsinf x x

' 1cos cot

sinf x x x

x

In interval 0,2

, ' cot 0f x x

f is strictly increasing in 0,2

.

In interval ,2

, ' cot 0f x x

f is strictly increasing in ,2

.

Question 17:



Prove that the function f is given by logcosf x x is strictly decreasing on 0,2

and

strictly increasing on ,2

Solution 17:

We have,

logcosf x x

' 1sin tan

cosf x x x

x

In interval 0,2

, tan 0 tan 0.x x

' 0 0,2

f x on

f is strictly decreasing on 0,2

.

In interval ,2

, tan 0 tan 0.x x

' 0 ,2

f x on

f is strictly increasing on ,2

.

Question 18:

Prove that the function given by 3 23 3 100f x x x x is increasing in R .

Solution 18:

We have,

3 23 3 100f x x x x

' 2

2

2

3 6 3

3 2 1

3 1

f x x x

x x

x

For any xR

(x – 1)2 > 0

Thus 'f x is always positive in R .

Hence, the given function f is increasing in R .

Question 19:

The interval in which 2 xy x e is increasing is



A. ,

B. 2,0

C. 2,

D. 0,2

Solution 19:

We have, 2 xy x e

22 2x x xdyxe x e xe x

dx

Now , 0.dy

dx

0 2x and x

The points 0x and 2x divided the real line into the three disjoint intervals

i.e., ,0 , 0,2 and 2, .

In intervals ,0 and 2, , ' 0 xf x as e is always positive.

f is decreasing on ,0 and 2, .

In interval 0,2 , ' 0f x .

f is strictly increasing on 0,2 .

Hence, f is strictly increasing in interval 0,2 .


Exercise 6.3

Question 1:

Find the slope of the tangent to the curve 43 4y x x at 4.x

Solution 1:

The given curve is 43 4y x x .

Then, the slope of the tangent to the given curve at 4x is given by,

33

44

12 4 12 4 4 12 64 4 764x

x

dxx

dy

.

Question 2:

Find the slope of the tangents to the curve 1

2

xy

x

, 2x at 10x .

Solution 2:



The given curve is 1

2

xy

x

.

2

2 2

2 1 1 1

2

2 1 1

2 2

x xdx

dy x

x x

x x

Thus, the slope of the tangent at 10x is given by,

2 2

1010

1 1 1.

642 10 2xx

dx

dy x

Hence, the slope of the tangent at 10x is 1

.64

Question 3:

Find the slope of the tangent to curve 3 1y x x at the point whose x-coordinate is 2.

Solution 3:

The given curve is 3 1y x x

23 1dx

xdy

The slope of the tangent to a curve at 0 0,x y is

0 0,x y

dx

dy

.

It is given that 0 2.x

Hence, the slope of the tangent at the point where the x-coordinate is 2 is given by,

22

22

3 1 3 2 1 12 1 11x

x

dxx

dy

Question 4:

Find the slope of the tangent to the curve 3 3 2y x x at the point whose x-coordinate is 3.

Solution 4:

The given curve is 3 3 2y x x

23 3dx

xdy

The slope of the tangent to a curve at 0 0,x y is

0 0,x y

dx

dy

Hence, the slope of the tangent at the tangent at the point where the x-coordinate is 3 is given

by,

22

33

3 3 3 3 3 27 3 24x

x

dxx

dy



Question 5:

Find the slope of the normal to the curve 3cosx a , 3siny a at .4

Solution 5:

It is given that 3cosx a and 3siny a .

2 2

2

2

2

3 cos sin 3 cos sin

3 sin cos

3 sin cos sintan

3 cos sin cos

dxa a

d

dya

d

dy

dx ad

dxdy a

d

Therefore, the slope of the tangent at 4

is given by,

4

4

tan tan 14

dx

dy

Hence, the slope of the normal at 4

is given by,

1 11

1slope of the tangent at

4

Question 6:

Find the slope of the normal to the curve 1 sinx a and 2cosy b at 2

.

Solution 6:

It is given, that 1 sinx a and 2y bco .

cosdx

ad

and 2 cos sin 2 sin cosdy

b bd

2 sin cos 2sin

cos

dy

dy b bd

dxdx a a

d

Therefore, the slope of the tangent at 2

is given by,



22

2 2 2sin sin

2

dy b b b

dx a a a

Hence, the slope of the normal at 2

is given by,

1 1

2 2slope of the tangent at

4

a

b b

a

Question 7:

Find the points at which tangent to the curve 3 23 9 7y x x x is parallel to the x - axis.

Solution 7:

The equation of the given curve is 3 23 9 7.y x x x

23 6 9dy

x xdx

Now, the tangent is parallel to the x - axis if the slope of the tangent is zero. 2 23 6 9 0 2 3 0x x x x

3 1 0

3 1

x x

x or x

When 3x , 3

3 9 3 7 27 27 27 7 20y .

When, 1x , 3 2

1 3 1 9 1 7 1 3 9 7 12.y

Hence, the points at which the tangent is parallel to the x -axis are 3, 20 and 1,12 .

Question 8:

Find a point on the curve 2

2y x at which the tangent is parallel to the chord joining the

points 2,0 and 4,4 .

Solution 8:

If a tangent is parallel to the chord joining the points 2,0 and 4,4 , then the slope of the

tangent = the slope of the chord.

The slope of the chord is 4 0 4

2.4 2 2

Now, the slope of the tangent to the given curve at a point ,x y is given by,

2 2dy

xdx

Since the slope of the tangent = slope of the chord, we have:



2

2 2 2

2 1 3

When 3, 3 2 1.

x

x x

x y

Hence, the required point is 3,1 .

Question 9:

Find the point on the curve 3 11 5y x x at which the tangent is 11.y x

Solution 9:

The equation of the given curve is 3 11 5y x x .

The equation of the tangent to the given curve is given as 11y x (which is of the form

y mx c ).

Slope of the tangent = 1

Now, the slope of the tangent to the given curve at the point ,x y is given by,

23 11dy

xdx

Then, we have: 2

2

2

3 11 1

3 12

4

2

x

x

x

x

When 3

2, 2 11 2 5 8 22 5 9.x y

When 3

2, 2 11 2 5 8 22 5 19.x y

Hence, the required points are 2, 9 and 2,19 .

Question 10:

Find the equation of all lines having slope –1 that are tangents to the curve 1

, 11

y xx

Solution 10:

The equation of the given curve is 1

, 11

y xx

The slope of the tangents to the given curve at any point ,x y is given by,

2

1

1

dy

dx x

If the slope of the tangents is –1, then we have:



2

2

11

1

1 1

1 1

2, 0

x

x

x

x

When 0x , 1y and when 2x , 1.y

Thus, there are two tangents to the given curve having slope –1. These are passing through the

points 0, 1 and (2, 1).

The equation of the tangent through (0, –1) is given by,

1 1 0

1

1 0

y x

y x

y x

The equation of the tangent through 2,1 , is given by,

1 1 2

1 2

3 0

y x

y x

y x

Hence, the equations of the required lines are 1 0y x and 3 0.y x

Question 11:

Find the equation of all lines having slope 2 which are tangents to the curve 1

, 3.3

y xx

Solution 11:


, 33

y xx

The slope of the tangent to the given curve at any point ,x y is given by,

2

1

3

dy

dx x

If the slope of the tangent is 2, then we have:

2

2

2

12

3

2 3 1

13

2

x

x

x

This is not possible since the L.H.S. is positive while the R.H.S. is negative.

Hence, there is no tangent to the given curve having slope 2.



Question 12:

Find the equations of all lines having slope 0 which are tangent to the curve 2

1.

2 3y

x x

Solution 12:


1.

2 3y

x x


2 2

2 2

2 2 2 1

2 3 2 3

x xdy

dx x x x x

If the slope of the tangent is 0, then we have:

22

2 10

2 3

2 1 0

1

x

x x

x

x

When 1 1

1, .1 2 3 2

x y

The equation of the tangent through 1

1,2

is given by,

1

0 12

10

2

y x

y

1

2y

Hence, the equation of the required line is 1

2y .

Question 13:

Find points on the curve 2 2

19 16

x y at which the tangents are

i) Parallel to x -axis ii) Parallel to y -axis

Solution 13:

The equation of the given curve is 2 2

19 16

x y

On differentiating both sides with respect to x , we have:



2 2. 0

9 16

16

9

x y dy

dx

dy x

dx y

(i) The tangent is parallel to the x -axis if the slope of the tangent is i.e.,16

0,9

x

y

which is

possible if 0.x

Then, 2 2

19 16

x y for 0x

2 16 4y y

Hence, the points at which the tangents are parallel to the x -axis are 0,4 and 0, 4 .

(ii) The tangent is parallel to the y -axis if the slope of the normal is 0, which gives

1 90 0.

1616

9

yy

xx

y

Then, 2 2

19 16

x y for 0.y

3x

Hence, the points at which the tangents are parallel to the y -axis are 3,0 and 3,0 .

Question 14:

Find the equations of the tangents and normal to the given curves at the indicated points:

I. 4 3 26 13 10 5 0,5y x x x x at

II. 4 3 26 13 10 5 1,3y x x x x at

III. 3 1,1y x at

IV. 2 0,0y x at

V. cos , sin4

x t y t at t

Solution 14:

(i). The equation of the curve is 4 3 26 13 10 5.y x x x x

on differentiating with respect to x , we get:

3 24 18 26 10dy

x x xdx

0,5

10dy

dx

Thus, the slope of the tangent at 0,5 is –10. The equation of the tangent id given as:

5 10 0y x

5 10

10 5

y x

x y



The slope of the normal at 0,5 is

1 1.

Slope of the tangent at 0,5 10

Therefore, the equation of the normal at 0,5 is given as:

1

5 010

y x

10 50

10 50 0

y x

x y

(ii) The equation of the curve is 4 3 26 13 10 5.y x x x x

On the differentiating with respect to x , we get:

3 24 18 26 10dy

x x xdx

1,3

4 18 26 10 2dy

dx

Thus, the slope of the tangent at 1,3 is 2. The equation of the tangent is given as:

3 2 1

3 2 2

2 1

y x

y x

y x


1 1.



1

3 12

2 6 1

2 7 0

y x

y x

x y

(iii) The equation of the curve is 3.y x

On differentiating with respect to x , we get:

2

2

1,1

3

3 1 3

dyx

dx

dy

dx

Thus, the slope of the tangent at 1,1 is 3 and the equation of the tangent is given as:

1 3 1

3 2

y x

y x


1 1.





1

1 13

3 3 1

3 4 0

y x

y x

x y

(iv) The equation of the curve is 2.y x


0,0

2

0

dyx

dx

dy

dx

Thus, the slope of the tangent at 0,0 is 0 and the equation of the tangent is given as:

0 0 0

0

y x

y


1 1,.


which is not defined.

Therefore, the equation of the normal at 0 0 0,0x y is given by

0 0.x x

(v) The equation of the curve is cos , sin .x t y t

cos , sin .x t y t

4

sin , cos

coscot

sin

cot 1t

dx dyt t

dt dt

dy

dy tdtt

dxdx t

dt

dyt

dx

The slope of the tangent at 4

t

is –1.

When 4

t

, 1

2x and

1

2y .

Thus, the equation of the tangent to the given curve at 4

t

i.e., 1 1

,2 2

is

1 11 .

2 2

1 10

2 2

2 0

y x

x y

x y



The slope of the normal at 4

t

is 1

1.

Slope of the tangent at4

t

Therefore, the equation of the normal to the given curve at 4

t

i.e., at 1 1

,2 2

is

1 11 .

2 2y x

x y

Question 15:

Find the equation of the tangent line to the curve 2 2 7y x x which is

(a) parallel to the line 2 9 0x y

(b) Perpendicular to the line 5 15 13.y x

Solution 15:

The equation of the given curve is 2 2 7y x x


2 2.dy

xdx

(a) The equation of the line is 2 9 0x y

2 9 0 , 2 9x y y x

This is of the form .y mx c

Slope of the line = 2

If a tangent is parallel to the line 2 9 0x y , then the slope of the tangent is equal to the slope

of the line.

Therefore, we have:

2 2 2

2 4

2

x

x

x

Now, 2x

4 4 7 7y

Thus, the equation of the equation of the tangent passing through 2,7 is given by,

7 2 2

2 3 0

y x

y x

Hence, the equation of the tangent line to the given curve (which is parallel to line 2 9 0)x y

is 2 3 0y x .

(b) The equation of the line is 5 15 13.y x

135 15 13, 3

5y x y x

This is form of .y mx c




If a tangent id perpendicular to the line 5 15 13y x , then the slope of the tangent is

1 1.

slope of the line 3

12 2

3

12 2

3

52

3

x

x

x

5

6x

Now, 5

6x

25 10 25 60 252 2177

36 6 36 36y

Thus, the equation of the tangent passing through 5 217

,6 36

is given by,

217 1 5

36 3 6

36 217 16 5

36 18

36 217 2 6 5

36 217 12 10

36 12 227 0

y x

yx

y x

y x

y x

Hence, the equation of the tangent line to the given curve (which is perpendicular to line

5 15 13y x ) is 36 12 227 0y x .

Question 16:

Show that the tangents to the curve 37 11y x at the points where 2x and 2x are

parallel.

Solution 16:

The equation of the given curve is 37 11y x .

221dy

xdx

The slope of the tangent to a curve at 0 0x y is 0 0,

.x y

dy

dx

Therefore, the slope of the tangent at the point where 2x is given by,



2

2

21 2 84x

dy

dx

It is observed that the slope of the tangents at the points where 2x and 2x are equal.

Hence, the two tangents are parallel.

Question 17:

Find the points on the curve 3y x at which the slope of the tangent is equal to the y -coordinate

of the point.

Solution 17:

The equation of the given curve is 3.y x

23dy

xdx

The slope of the tangent at the point ,x y is given by,

2

,

3x y

dyx

dx

When the slope of the tangent is equal to the y-coordinate of the point, then 23 .y x

Also, we have 3.y x

2 3

2

3

3 0

0, 3

x x

x x

x x

When 0x , then 0y and when 3x then 2

3 3 27.y

Hence, the required points are 0,0 and 3,27 .

Question 18:

For the curve 3 54 2 ,y x x find all the points at which the tangents passes through the origin.

Solution 18:

The equation of the given curve is 3 54 2 .y x x

2 412 10dy

x xdx

Therefore, the slope of the tangent at a point ,x y is 2 412 10 .x x

The equation of the tangent at ,x y is given by,

2 412 10Y y x x X x ….. (1)

When the tangent passes through the origin 0,0 , then 0.X Y

Therefore, equation (1) reduces to:



2 4

3 5

12 10

12 10

y x x x

y x x

Also, we have 3 54 2 .y x x

3 5 3 5

5 3

5 3

3 2

12 10 4 2

8 8 0

0

1 0

0, 1

x x x x

x x

x x

x x

x

When 3 5

0, 4 0 2 0 0.x y

When 3 5

1, 4 1 2 1 2.x y

When 3 5

1, 4 1 2 1 2x y .

Hence, the required points are 0,0 , 1,2 and 1, 2 .

Question 19:

Find the points on the curve 2 2 2 3 0x y x at which the tangents are parallel to the x -axis.

Solution 19:

The equation of the given curve is 2 2 2 3 0x y x .

On differentiating with respect to x , we have:

2 2 2 0

1

1

dyx y

dx

dyy x

dx

dy x

dx y

Now, the tangents are parallel to the x -axis if the slope of the tangent is 0.

10 1 0 1

xx x

y

But, 2 2 2 3 0x y x for 1.x

2 4 , 2y y

Hence, the points at which the tangents are parallel to the x -axis are 1,2 and 1, 2 .

Question 20:

Find the equation of the normal at the point 2 3,am am for the curve 2 3.ay x

Solution 20:



The equation of the given curve is 2 3.ay x

On differentiating with respect to x , we have:

2

2

2 3

3

2

dyay x

dx

dy x

dx ay

The slope of a tangent to the curve at 0 0,x y is 0 0,x y

dy

dx

.

The slope of the tangent to the given curve at 2 3,am am is

2 3

22 2 4

2 33,

3 3 3.

2 22am am

amdy a m m

dx a ma am

Slope of normal at 2 3,am am

= 2 3

1 2

3slope of the tangent at , mam am

Hence, the equation of the normal at 2 3,am am is given by,

3 2

4 2

2 2

2

3

3 3 2 2

2 3 2 3 0

y am x amm

my am x am

x my am m

Question 21:

Find the equation of the normal to the curve 3 2 6y x x which are parallel to the line

14 4 0.x y

Solution 21:

The equation of the given curve is 3 2 6y x x .


23 2dy

xdx

Slope of the normal to the given curve at any point ,x y

=

1

Slope of the tangent at the point ,x y

2

1

3 2x

The equation of the given line is 14 4 0.x y

14 4 0x y , 1 4

14 14y x (which is of the form y mx c )



Slope of the given line = 1

14

If the normal is parallel to the line, then we must have the slope of the normal being equal to the

slope of the line.

2

2

2

2

1 1

3 2 14

3 2 14

3 12

4

2

x

x

x

x

x

When 2, 8 4 6 18.x y

When 2, 8 4 6 6.x y

Therefore, there are two normal to the given curve with slope 1

14

and passing through the points

2,18 and 2, 6 .

Thus, the equation of the normal through 2,18 is given by,

1

18 214

14 252 2

14 254 0

y x

y x

x y

And, the equation of the normal through 2, 6 is given by,

16 2

14

16 2

14

14 84 2

14 86 0

y x

y x

y x

x y

Hence, the equations of the normal to the given curve (which are parallel to the given line) are

14 254 0x y and 14 86 0.x y

Question 22:

Find the equations of the tangent and normal to the parabola 2 4y ax at the point 2 ,2at at .

Solution 22:

The equation of the given parabola is 2 4y ax .

On differentiating 2 4y ax with respect to x , we have:



2 4

2

dyy a

dx

dy a

dx y

The slope of the tangent at 2 ,2at at is

2 ,2

2 1.

2at at

dy a

dx at t

Then, the equation of the tangent at 2 ,2at at is given by,

2

2 2

2

12

2

y at x att

ty at x at

ty x at

Now, the slope of the normal at 2 ,2at at is given by,

2

1

Slope of the tangent ,2t

at at

Thus, the equation of the normal at 2 ,2at at is given as:

2

3

3

2

2

2

y at t x at

y at tx at

y tx at at

Question 23:

Prove that the curves 2x y and xy k cut at angles if 28 1.k

[Hint: Two curves intersect at right angle if the tangents to the curves at the point of intersection

are perpendicular to each other.]

Solution 23:

The equation of the given curves are given as 2x y and xy k

Putting 2x y in xy k , we get:

1

3 3y k y k 2

3x k

Thus, the point of intersection of the given curves is 2 1

3 3,k k

Differentiating 2x y with respect to x , we have:

11 2

2

dy dyy

dx dx y



Therefore, the slope of the tangent to the curve 2x y at 2 1

3 3,k k

is 2 1

3 3

1

, 3

1

2k k

dy

dxk

On differentiating xy k with respect to x , we have:

0dy dy y

x ydx dx x

Slope of the tangent to the curve xy k at 2 1

3 3,k k

is given by,

2 1 2 1

3 3 3 3

1

3

2 1

, , 3 3

1

k k k k

dy y k

dx xk k

We know that two curves intersect at right angles if the tangents to the curves at the point of

intersection i.e., at 2 1

3 3,k k

are perpendicular to each other.

This implies that we should have the product of the tangents as –1.

Thus, the given two curves cut at right angles if the product of the slopes of their respective

tangents at 2 1

3 3,k k

is –1.

i.e., 1 1

3 3

1 11

2k k

2

3

32

33

2

2 1

2 1

8 1

k

k

k

Hence, the given two curves cut at right angles if 28 1k .

Question 24:

Find the equations of the tangent and normal to the hyperbola 2 2

2 21

x y

a b at the point 0 0x y .

Solution 24:

Differentiating 2 2

2 21

x y

a b with respect to x , we have:



2 2

2 2

2

2

2 20

2 2

x y dy

a b dx

y dy x

b dx a

dy b x

dx a y

Therefore, the slope of the tangent at 0 0,x y is 0 0

2

0

2

, 0x y

b xdy

dx a y

Then, the equation of the tangent at 0 0,x y is given by,

2

00 02

0

2 22 2 2 2

0 0 0 0

2 22 2 2 2

0 0 0 0 0

b xy y x x

a y

a yy a y b xx b x

b xx a yy b x a y

2 2

0 0 0 0

2 2 2 20

xx yy x y

a b a b

2 2on dividing both sides by a b

0 0

2 21 0

xx yy

a b

2 2

0 0 2 2, lies on the hyperbola 1

x yx y

a b

0 0

2 21

xx yy

a b

Now, the slope of the normal at 0 0,x y is given by,

2

0

2

0 0 0

1

Slope of the tangent ,

a y

x y b x

Hence, the equation of the normal at 0 0,x y is given by,

2

00 02

0

00

2 2

0 0

00

2 2

0 0

0

a yy y x x

b x

x xy y

a y b x

x xy y

a y b x

Question 25:

Find the equation of the tangent to the curve 3 2y x which is parallel to the line

4 2 5 0x y .

Solution 25:





3

2 3 2

dy

dx x

The equation of the given line is 4 2 5 0x y .

4 2 5 0x y , 5

22

y x (which is of the form y mx c )


Now, the tangent to the given curve is parallel to the line 4 2 5 0x y if the slope of the

tangent is equal to the slope of the line.

32

2 3 2

33 2

4

93 2

16

9 413 2

16 16

41

48

x

x

x

x

x

When 41 41 41 41 32 9 3

, 3 2 2 .48 48 16 16 16 4

x y

Equation of the tangent passing through the point 41 3

,48 4

is given by,

3 412

4 48

4 3 48 412

4 48

48 414 3

6

24 18 48 41

y x

y x

xy

y x

48 24 23x y

Hence, the equation of the required tangent is 48 24 23x y .

Question 26:

The slope of the normal to the curve 22 3siny x x at 0x is

(A) 3 , (B) 1

3, (C) 3 , (D)

1

3

Solution 26:



The equation of the given curve is 22 3siny x x .

Slope of the tangent to the given curve at 0x is given by,

0

0

4 3cos 0 3cos0 3x

x

dyx x

dx

Hence, the slope of the normal to the given curve at 0x is

1 1.

Slope of the tangent a 0 3t x


Question 27:

The line 1y x is a tangent to the curve 2 4y x at the point

(A) 1,2 , (B) 2,1 , (C) 1, 2 , (D) 1,2

Solution 27:


Differentiating with respect to x , we have:

22 4

dy dyy

dx dx y

Therefore, the slope of the tangent to the given curve at any point ,x y is given by,

2dy

dx y

The given line is 1y x ( which is of the form y mx c )

Slope of the line = 1.

The line 1y x is a tangent to the given curve if the slope of the line is equal to the slope of

the tangent. Also, the line must intersect the curve.

Thus, we must have:

21

2

y

y

Now, 1 1 2 1 1y x x y x

Hence, the line 1y x is a tangent to the given curve at the point 1,2 .

The correct answer is A.

Exercise 6.4

Question 1:

Using differentials, find the approximate value of each of the following up to 3 places of decimal.



(i) 25.3 , (ii) 49.5 , (iii) 0.6 , (iv) 1

30.009 , (v) 1

100.999 , (vi) 1

415 , (vii) 1

326 ,

(viii) 1

4255 , (ix) 1

482 , (x) 1

2401 , (xi) 1

20.0037 , (xii) 1

326.57 , (xiii) 1

481.5 ,

(xiv) 3

23.968 , (xv) 1

532.15

Solution 1:

(i) 25.3

Consider y x . Let 25x and 0.3.x

Then,

25.3 25 25.3 5y x x x

25.3 5y

Now, dy is approximately equal to y and is given by,

1

0.32

dydy x

dx x

as y x

1

0.3 0.032 25

Hence, the approximate value of 25.3 is 0.03 5 5.03.

(ii) 49.5


Then,

49.5 49 49.5 7y x x x

49.5 7 y


1

0.52

dydy x

dx x

as y x

1 1

0.5 0.5 0.035142 49

Hence, the approximate value of 49.5 is 7 0.035 7.035.

(iii) 0.6


Then,

0.6 1y x x x

0.6 1 y


1

2

dydy x x

dx x

as y x



1

0.4 0.22

Hence, the approximate value of 0.6 is 1 0.2 1 0.2 0.8.

(iv) 1

30.009

Consider

1

3y x . Let 0.008x and 0.001.x

Then,

1 1 1 1 1

3 3 3 3 3

1

3

0.009 0.008 0.009 0.2

0.009 0.2

y x x x

y


2

3

1

3

dydy x x

dxx

1

3as y x

1 0.001

0.001 0.0083x0.04 0.12

Hence, the approximate value of 1

30.009 is 0.2 0.008 0.208.

(v) 1

100.999

Consider 1

10y x . Let 1x and 0.001.x

Then,

1 1 1

10 10 100.999 1y x x x

1

100.999 1 y


9

10

1

10

dydy x x

dxx

1

10as y x

1

0.001 0.000110


100.999 is 1 0.0001 0.9999.

(vi) 1

415

Consider

1

4y x . Let 16x and 1.x

Then,

11 1 1 144 4 4 415 16 15 2y x x x

14

15 2 y




3

4

1

4

dydy x x

dxx

1

4as y x

3

4

1 1 11 0.03125

4x8 324 16


415 is 2 0.03125 1.96875.

(vii) 1

326

Consider 1


1 1 1 1 1

3 3 3 3 326 27 26 3y x x x

1

326 3 y


2

3

1

3

dydy x x

dxx

1

3as y x

2

3

1 11 0.0370

273 27


326 is 3 0.0370 2.9629.

(viii) 1

4255

Consider 1


Then,

1 1 1 1 1

4 4 4 4 4255 256 255 4y x x x

1

4255 4 y


3

4

1

4

dydy x x

dxx

1

4as y x

3 3

4

1 11 0.0039

4x44 256


4255 is 4 0.0039 3.9961.

(iX) 1

482



Consider

1


1 1 1 1 1

4 4 4 4 482 81 82 3y x x x

1

482 3y


3

4

1

4

dydy x x

dxx

1

4as y x

3 3

4

1 1 11 0.009

1084 34 81


482 is 3 0.009 3.009.

(x) 1

2401

Consider

1


401 400 401 20y x x x

401 20 y


1

2

dydy x x

dx x

1

2as y x

1 1

1 0.0252x20 40

Hence, the approximate value of 401 is 20 0.025 20.025 .

(xi) 1

20.0037

Consider

1

2y x . Let 0.0036x and 0.0001.x

Then,

1 1 1 1 1

2 2 2 2 20.0037 0.0036 0.0037 0.06y x x x

1

20.0037 0.06 y


1

2

dydy x x

dx x

1

2as y x

1

0.00012x0.06

0.00010.00083

0.12



Thus, the approximately value of 1

20.0037 is 0.06 0.00083 0.6083.

(xii) 1

326.57

Consider

1

3y x . Let 27x and 0.43.x

Then,

11 1 1 133 3 3 326.57 27 26.57 3y x x x

1

326.57 3 y


2

3

1

3

dydy x x

dxx

1

3as y x

10.43

3 9

0.430.015

27


326.57 is 3 0.015 2.984.

(xiii) 1

481.5

Consider

1


Then,

1 1 1 1 1

4 4 4 4 481.5 81 81.5 3y x x x

1

481.5 3 y


3

4

1

4

dydy x x

dxx

1

4as y x

3

1 0.50.5 0.0046

1084 3


481.5 is 3 0.0046 3.0046.

(xiv) 3

23.968

Consider,

3

2y x .Let 4x and 0.032.x

Then,

33 3 3 322 2 2 23.968 4 3.968 8y x x x



3

23.968 8 y


1

23

2

dydy x x x

dx

3

2as y x

3

2 0.0322

0.096


23.968 is 8 0.096 7.904.

(XV) 1

1532.15

Consider

1


Then,

11 1 1 155 5 5 532.15 32 32.15 2y x x x

1

1532.15 2 y


4

5

1.

5

dydy x x

dxx

1

5as y x

4

10.15

5x 2

0.150.00187

80


1532.15 is 2 0.00187 2.00187.

Question 2:

Find the approximate value of (2.01), where 24 5 2f x x x .

Solution 2:

Let 2x and 0.01.x Then, we have:

2

2.01 4 5 2f f x x x x x x

Now, y f x x f x

f x x f x y

' .f x f x x as dx x

22.01 4 5 2 8 5f x x x x




2

4 2 5 2 2 8 2 5 0.01 2, 0.01as x x

16 10 2 16 5 0.01

28 21 0.01

28 0.21

28.21

Hence, the approximate value of 2.01f is 28.21.

Question 3:

Find the approximate value of 5.001f , where 3 27 15.f x x x

Solution 3:

Let 5x and 0.001x . Then, we have:

3 2

5.001 7 15f f x x x x x x

Now, y f x x f x

' .

f x x f x y

f x f x x

as dx x

3 2 25.001 7 15 3 14f x x x x x

3 2 2

5 7 5 15 3 5 14 5 0.001 5, 0.001x x

125 175 15 75 70 0.001

35 5 0.001

35 0.005

34.995

Hence, the approximate value of 5.001f is 34.995.

Question 4:

Find the approximate change in the volume V of a cube side x meters caused by increasing side

by 1%.

Solution 4:

The volume of a cube V of side x is given by 3.V x

dVdV x

dx

23x x

23 0.01x x 1% 0.01as of x is x

30.03x



Hence, the approximate change in the volume of the cube is 3 30.03 m .x

Question 5:

Find the approximate change in the surface area of a cube of side x meters caused by decreasing

the side by 1%.

Solution 5:

The surface area of a cube S of side x is given by 26 .S x

ds dsx

dx dx

12x x

12 0.01x x 1% 0.01as of x is x

20.12x

Hence, the approximate change in the surface area of the cube is 2 20.12 m .x

Question 6:

If the radius of a sphere is measured as 7 m with an error of 0.02 m, then find the approximate

error in calculating its volume.

Solution 6:

Let r be the radius of the sphere and r be the error in measuring the radius.

Then,

7r m and 0.02r m

Now, the volume V of the sphere is given by,

3

2

4

3

4

V r

dVr

dr

dVdV r

dr

24 r r

2 3 34 7 0.02 3.92m m

Hence, the approximate error in calculating the volume is 3.92 m3.

Question 7:

If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate

error in calculating in surface area.



Solution 7:

Let r be the radius of the sphere and r be the error in measuring the radius.

Then,

9r m and 0.03r m

Now, the surface area of the sphere S is given by,

8dS

rdr

dSdS r

dr

2

2

8

8 9 0.03

2.16

r r

m

m

Hence, the approximate error in calculating the surface area is 2.16 m2.

Question 8:

If 23 15 5f x x x , then the approximate value of (3.02) is

(A) 47.66 , (B) 57.66 , (C) 67.66 , (D) 77.66

Solution 8:

Let 3x and 0.02x Then, we have:

2

3.02 3 15 5f f x x x x x x

Now, y f x x f x

f x x f x y

'f x f x x As dx x

23.02 3 15 5 6 15f x x x x

23 3 15 3 5 6 3 15 0.02 3, 0.02As x x

27 45 5 18 15 0.02

77 33 0.02

77 0.66

77.66

Hence, the approximate value of (3.02) is 77.66.


Question 9:

The approximate change in the volume of a cube of side x meters caused by increasing the side

by 3% is

A. 3 30.06x m B. 3 30.6x m C. 3 30.09x m D. 3 30.9x m



Solution 9:

The volume of a cube V of side x is given by 3.V x

dVdV x

dx

2

2

3

3 0.03

x x

x x

3% 0.03As of axis x

3 30.09x m

Hence, the approximate change in the volume of the cube is 30.09x m3.

The correct answer is C.

Exercise 6.5

Question 1:

Find the maximum and minimum values, if any, of the following given by

(i) 2

2 1 3f x x (ii) 29 12 2f x x x

(iii) 2

1 10f x x (iv) 3 1g x x

Solution 1:

(i) The given function is 2

2 1 3f x x

It can be observed that 2

2 1 0x for every .xR

Therefore, 2

2 1 3 3f x x for every .xR

The minimum value of f is attained when 2 1 0.x

12 1 0 ,

2x x

Minimum value of

21 1

2. 1 3 3.2 2

f

Hence, function f does not have a maximum value.

(ii) The given function is 2

2 29 12 2 3 2 2.f x x x x


23 2 0x for every .xR

Therefore, 2

23 2 2 2f x x for every .xR

The minimum value of f is attained when 3 2 0.x

23 2 0 0 ,

3x x



Minimum value of

2

2 23 2 2 2

3 3f


(iii) The given function is 2

1 10f x x .


1 0x for every .xR

Therefore, 2

1 10 10f x x for every .xR

The maximum value of f is attained when 1 0.x

1 0 , 0x x

Maximum value of 2

1 1 1 10 10f f


(iv) The given function is 3 1g x x .

Hence, function g neither has a maximum value nor a minimum value.

Question 2:

Find the maximum and minimum values, if any, of the following functions given by

(i) 2 1f x x (ii) 1 3g x x (iii) sin 2 5h x x

(iv) sin 4 3f x x (v) 4, 1,1h x x x

Solution 2:

(i) 2 1f x x

We know that 2 0x for every .xR

Therefore, 2 1 1f x x for every .xR

The minimum value of f is attained when 2 0x .

2 0

2

x

x

Minimum value of 2 2 2 1 1f f


(ii) 1 3g x x

We know that 1 0x for every .xR

Therefore, 1 3 3g x x for every .xR

The maximum value of g is attained when 1 0x .

1 0

1

x

x

Maximum value of 1 1 1 3 3g g

Hence, function g does not have a maximum value.



(iii) sin2 5h x x

We know that 1 sin2 1.x

1 5 sin 2 5 1 5

4 sin 2 5 6

x

x

Hence, the maximum and minimum values of h are 6 and 4 respectively.

(iv) sin 4 3f x x

We know that 1 sin4 1.x

2 sin 4 3 4

2 sin 4 3 4

x

x

Hence, the maximum and minimum values of f are 4 and 2 respectively.

(v) 4, 1,1h x x x

Here, if a point 0

x is closest to 1 , then we find 0

01 1

2

xx for all

01,1x .

Also if 1

x is closet to 1 , then we find 11

11 1

2

xx

for all

01,1x .

Hence, function h x has neither maximum nor minimum value in 1,1 .

Question 3:

Find the local maxima and local minima, if any, of the following functions. Find also the local

maximum and the local minimum values, as the case may be:

(i) 2f x x (ii) 3 3g x x x (iii) sin cos.02

h x x x

(iv) sin cos ,0 2f x x x x (v) 3 26 9 15f x x x x

(vi) 2

, 02

xg x x

x (vii) 2

1

2g x

x

(vii) 1 , 0f x x x x

Solution 3:

(i) 2f x x

' 2f x x

Now,

' 0 0f x x

Thus, 0x is the only critical point which could possibly be the point of local maxima or local

minima of f .

We have f’’(0)=2, which is positive.

Therefore, by second derivative test, 0x is a point of local minima and local minimum value

of f

at 0x is 0 0.f

(ii) 3 3g x x x

' 23 3g x x



Now,

2' 0 3 3 1g x x x

g’’(x) = 6x

g’’(1) = 6 > 0

g’’(–1) = –6 < 0

By second derivative test, 1x is a point of local minima and local minimum value of g

At 1x is 31 1 3 1 3 2.g However,

1x is a point of local maxima and local maximum value of g at

1x is 3

1 1 3 1 1 3 2.g

(iii) sin cos.02

h x x x

'( ) cos sin

'( ) 0 sin cos tan 1 0,4 2

h x x x

h x x x x x

h’’(x) = –sin x – cos x

1 1 1'' 2 0

4 2 2 2h

Therefore, by second derivative test, 4

x

is a point of local maxima and the local Maximum

value of h at 4

x

is 1 1

sin cos 2.4 4 4 2 2

h

(iv) sin cos ,0 2f x x x x

'

'

cos sin

3 70 cos sin tan 1 , 0,2

4 4

'' sin cos

3 3 3 1 1'' sin cos 2 0

4 4 4 2 2

7 7 7 1 1'' sin cos 2 0

4 4 4 2 2

f x x x

f x x x x x

f x x x

f

f


4x

is appoint of local maxima and the local maximum

value of f at 3

4x

is

3 3 3 1 1sin cos 2.

4 4 4 2 2f

However,

7

4x

is a point of local minima and the

local minimum value of f at 7

4x

is

7 7 7 1 1sin cos 2.

4 4 4 2 2f

(v) 3 6 9 15f x x x x



' 2

2

3 12 9

0 3 4 3 0

3 1 3 0

1,3

f x x x

f x x x

x x

x

Now, "f

"

"

6 12 6 2

1 6 1 2 6 0

3 6 3 2 6 0

x x x

f

f

Therefore, by second derivative test, 1x is a point of local maxima and the local maximum

value of f at 1x is 1 1 6 9 15 19.f However, 3x is a point of local minima and

the local minimum value of f at 3x is 3 27 54 27 15 15.f

(vi) 2

, 02

xg x x

x

'

2

1 2

2g x

x

Now,

' 3

2

2 10 gives 4 2

2g x x x

x

Since 0x , we take 2.x

Now,

3

3

4''

4 1'' 2 0

2 2

g xx

g

Therefore, by second derivative test, 2x is a point of local minima and the local minimum

value of g at 2x is 2 2

2 1 1 2.2 2

g

(vii) 2

1

2g x

x

'

23

'

23

(2 )

2

20 0 0

2

xg x

x

xg x x

x

Now , for values close to 0x and to the left of 0 , 1 0.g x Also, for values close to 0x

and to the right of 1 0.g x

Therefore, by first derivative test 0x is a point of local maxima and the local maximum value

of 0g is1 1

.0 2 2



(viii) 1 , 0f x x x x

3

2

1'( ) 1 ( 1) 1

2 1 2 1

2(1 ) 2 3

2 1 2 1

2 3 2'( ) 0 0 2 3 0

32 1

11 ( 3) (2 3 )

1 2 1''( )

2 1

11 ( 3) 2(2 3 )

2 1

2(1 )

6(1 ) 2(2 3 )

4(1 )

3 4

4(1

xf x x x x x

x x

x x x

x x

xf x x x

x

x xx

f xx

x xx

x

x x

x

x

3

2)x

3 3 3

2 2 2

23 4

2 2 4 13' 0

32 1 1

4 1 4 23 3 3

f


3x is a point of local maxima and the local maximum

value of f at 2

3x is

2 2 2 2 1 2 2 31 .

3 3 3 3 3 93 3f

Question 4:

Prove that the following functions do not have maxima or minima:

(i) xf x e (ii) logg x x (iii) 3 2 1h x x x x

Solution 4:

(i) xf x e

' xf x e



Now, if ' 0,f x then 0xe . But the exponential function can never assume 0 for any value

of .x

Therefore, there does not exist cR such that ' 0.f c

Hence, function f does not have maxima or minima.

(ii) We have,

logg x x

' 1g x

x

Since log x is defined for a positive number x , ' 0g x for any .x

Therefore, there does not exist cR such that ' 0g c .

Hence, function g does not have maxima or minima.

(iii) We have,

3 2 1h x x x x

' 23 2 1h x x x

Now,

Therefore, there does not exist cR such that ' 0.h c

Hence, function h does not have maxima or minima.

Question 5:

Find the absolute maximum value and the absolute minimum value of the following functions

in the given intervals:

(i) 3, 2,2f x x x (ii) sin cos , 0,f x x x x

(iii) 21 94 , 2,

2 2f x x x x

(iv)

21 3, 3,1f x x x

Solution 5:

(i) The given function is 3f x x .

' 23f x x

Now,

' 0 0f x x

Then, we evaluate the value of f at critical point 0x and at end points of the interval 2,2 .

3

3

0 0

2 2 8

2 2 8

f

f

f

Hence, we can conclude that the absolute maximum value of f on 2,2 is 8 occurring at

2.x Also, the absolute minimum value of f on 2,2 is 8 occurring at 2.x



(ii) The given function is sin cosf x x x .

' cos sinf x x x

Now,

' 0 sin cos tan 14

f x x x x x

Then, we evaluate the value of f at critical point 4

x

and at the end points of the interval

0, .

1 1 2sin cos 2

4 4 4 2 2 2

0 sin 0 cos0 0 1 1

sin cos 0 1 1

f

f

f

Hence, we can conclude that the absolute maximum value of f on 0, is 2 occurring at

4x

and the absolute minimum value of f on 0, is 1 occurring .x

(iii) The given function is 214

2f x x x

' 14 2 4

2f x x x x

Now,

' 0 4f x x

Then, we evaluate the value of f at critical point 4,x and at the end points of the interval

92,

2

.

1

4 16 16 16 8 82

f

2

12 8 4 8 2 10

2

9 9 1 9 814 18 18 10.125 7.875

2 2 2 2 8

f

f

Hence, we can conclude that the absolute maximum value of f on 9

2,2

is 8 occurring at

4x and the absolute minimum value of f on 9

2,2

is 10 occurring at 2.x

(iv) The given function is 2

1 3f x x .

' 2 1f x x

Now,

' 0 2 1 0, 1f x x x



Then, we evaluate the value of f at critical point 1x and at the end points of the interval

3,1 .

2

2

1 1 1 3 0 3 3

3 3 1 3 16 3 19

f

f

Hence, we can conclude that the absolute maximum value of f on 3,1 is 19 occurring at

3x and the minimum value of f on 3,1 is occurring at 1x .

Question 6:

Find the maximum profit that a company can make, if the profit function is given by

241 24 18p x x x

Solution 6:

The profit function is given as 241 24 18p x x x .

'

''

24 36

36

p x x

p x

Now,

' 24 20

36 3p x x

Also,

2'' 36 0

3p

By second derivatives test, 2

3x is the point of local maximum of p .

Maximum profit = p(-2/3)2

2 241 24 18

3 3

41 16 8

49

Hence, the maximum profit that the company can make is 49 units.

Question 7:

Find the intervals in which the function f given by 3

3

1, 0f x x x

x is

(i) Increasing (ii) Decreasing

Solution 7:

3

3

1f x x

x



6

' 2

4 4

3 3 33

xf x x

x x

Then, ' 6 60 3 3 0 1 1f x x x x

Now, the points 1x and 1x divided the real line into three disjoint intervals i.e.,

, 1 , 1,1 and 1, .

In intervals , 1 and 1, i.e., when 1x and 1,x ' 0.f x

Thus, when 1x and 1,x f is increasing.

In intervals 1,1 i.e., when 1 1,x ' 0.f x

Thus, when 1 1,x f is decreasing.

Question 8:

At what points in the interval 0,2 , does the function sin2x attain, its maximum value?

Solution 8:

Let sin2 .f x x

' 2cos2f x x

Now,

' 0 cos 2 0

3 5 72 , , ,

2 2 2 2

3 5 7, , ,

4 4 4 4

f x x

x

x

Then, we evaluate the values of f at critical points 3 5 7

, , ,4 4 4 4

x

and at the end points of

the interval 0,2 .

3 3sin 1, sin 1

4 2 4 2

5 5 7 7sin 1, sin 1

4 2 4 2

0 sin 0 0, 2 sin 2 0

f f

f f

f f

Hence, we can conclude that the absolute maximum value of f 0,2 is occurring at4

x

and 5

4x

.

Question 9:

What is the maximum value of the function sin cosx x ?



Solution 9:

Let sin cosf x x x

'

'

'

cos sin

50 sin cos tan 1 , ...,

4 4

sin cos sin cos

f x x x

f x x x x x

f x x x x x

Now, ''f x will be negative when sin cosx x is positive i.e., when sin x and cos x are

both positive. Also, we know that sin x and cos x both are positive in the first quadrant. Then,

''f x will be negative when 0,2

x

.

Thus, we consider 4

x

.

" 2sin cos 2 0

4 4 4 2f

By second derivative test, f will be the maximum at 4

x

and the maximum value of f is

1 1 2sin cos x 2

4 4 4 2 2 2f

.

Question 10:

Find the maximum value of 32 24 107x x in the interval 1,3 . Find the maximum value of

the same function in 3, 1 .

Solution 10:

Let 32 24 107f x x x

' 2 26 24 6 4f x x x

Now,

' 2 20 6 4 0 4 2f x x x x

We first consider the interval 1,3 .

Then, we evalutat the value of f at the critical point 1,3x and at the end points of the interval

[1, 3]

2 2 8 24 2 107 16 48 107 75

1 2 1 24 1 107 2 24 107 85

3 2 27 24 3 107 54 72 107 89

f

f

f

Hence, the absolute maximum value of f x in the interval 1,3 is 89 occurring at 3.x

Next, we consider the interval 3, 1 .



Evaluate the value of f at the critical pont 2 [ 3, 1]x and at the end points of the interval

[1, 3].

3 2 27 24 3 107 54 72 107 125

1 2 1 24 1 107 2 24 107 129

2 2 8 24 2 107 16 48 107 139

f

f

f

Hence, the absolute maximum value of f x in the interval 3, 1 is 139occurring at 2.x

Question 11:

It is given that at 1x , the function 4 262 9x x ax attains its maximum value, on the interval

0,2 . Find the value of a .

Solution 11:

Let 4 262 9f x x x ax .

' 24 124f x x x a

It is given that function f attains its maximum value on the interval 0,2 at 1x .

' 1 0

4 124 0

120

f

a

a

Hence, the value of a is 120.

Question 12:

Find the maximum and minimum values of sin2x x on 0,2 .

Solution 12:

Let sin 2f x x x .

' 1 2cos2f x x

Now, ' 1 20 cos2 cos cos cos

2 3 3 3f x x

'

22 2

3

,3

2 4 5, , , 0,2

3 3 3 3

x n

x n

x

Z

Z

Then, we evaluate the value of f at critical points 2 4 5

, , ,3 3 3 3

x

and the end points of

the interval 0,2 .



2 3sin

3 3 3 3 2

2 2 4 2 3sin

3 3 3 3 2

4 4 8 4 3sin

3 3 3 3 2

5 5 10 5 3sin

3 3 3 3 2

0 0 sin 0 0

2 2 sin 4 2 0 2

f

f

f

f

f

f

Hence, we can conclude that the absolute maximum value of f x in the interval 0,2 is 2

occurring at 2x and the absolute minimum value of f x in the interval 0,2 is 0

occurring at 0x .

Question 13:

Find two numbers whose sum is 24 and whose product is as large as possible.

Solution 13:

Let one number be x . Then, the other number is 24 .x

Let p x denote the product of the two numbers. Thus, we have:

2

'

"

24 24

24 2

2

P x x x x x

P x x

P x

Now,

' 0 12P x x

Also,

" 12 2 0P

By second derivative test, 12x is the point of local maxima of P . Hence, the product of the

numbers is the maximum when the numbers are 12 and 24 12 12.

Question 14:

Find two positive numbers x and y such that 60x y and 3xy is maximum.

Solution 14:

The two numbers are x and y such that 60x y .

60y x

Let 3f x xy



3

3 2'

60

60 3 60

f x x x

f x x x x

3

3

60 60 3

60 60 4

x x x

x x

And, 2" 2 60 60 4 4 60f x x x x

2 60 60 4 2 60

2 60 180 6

12 60 30

x x x

x x

x x

Now, ' 0 60 or 15f x x x

When 60x , " 0.f x

When, 15x , '' 12 60 15 30 15 12 45 15 0.f x

By second derivative test, 15x is a point of local maxima of f . Thus , function 3xy is

maximum when 15x and 60 15 45.y

Hence, the required numbers are 15 and 45.

Question 15:

Find two positive numbers x and y such that their sum is 35 and the product 2 5x y is a

maximum.

Solution 15:

Let one number be x . Then, the other number is 35 .y x

Let 2 5p x x y . Then, we have:

52

5 4' 2

35

2 35 5 35

P x x x

P x x x x x

4

35 2 35 5x x x x

4

4

35 70 7

7 35 10

x x x

x x x

And, 4 4 3" 7 35 10 7 35 5 4 35 10P x x x x x x

4 4 3

3

3 2 2 2

3 2

7 35 10 7 35 28 35 10

7 35 35 10 35 4 10

7 35 350 45 35 40 4

7 35 6 120 350

x x x x x x x

x x x x x x x

x x x x x x x

x x x



Now, ' 0 0, 35, 10P x x x x

When, 35,x ' 0f x f x and 35 35 0y . This will make the product 2 5x y equal to

0 .

When 0, 35 0 35x y and the product 2 2x y will be 0 .

0x and 35x cannot be the possible values of x .

When 10,x we have:

3" 7 35 10 6x100-120x10+350P x

3

7 25 250 0

By second derivative test, P x will be the maximum when 10x and 35 10 25y .

Hence, the required numbers are 10 and 25.

Question 16:

Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Solution 16:

Let one number be .x Then, the other number is 16 .x

Let the sum of the cubes of these numbers be denoted by .S x Then,

33

2' 2 "

16

3 3 16 , 6 6 16

S x x x

S x x x S x x x

Now, 2' 20 3 3 16 0S x x x

22

2 2

16 0

256 32 0

2568

32

x x

x x x

x

Now, " 8 6 8 6 16 8 48 48 96 0S

By second derivative test, x = 8 is the point of local minima of S.

Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and 16 –8

= 8.

Question 17:

A square piece of tin od side 18 cm is to made into a box without top, by cutting a square from

each corner and folding up the flaps to form the box. What should be the side of the square to

be cut off so that the volume of the box is the maximum possible?

Solution 17:

Let the side of the square to be cut off be x cm.Then, the length and the breath of the box will

be (18 – 2x) cm each and the height of the box is x cm

Therefore, the volume V(x) of the box is given by,



2( ) (18 2 )V x x x

2'( ) (18 2 ) 4 (18 2 )V x x x x

(18 2 )[18 2 4 ]

(18 2 )[18 6 ]

6 2(9 )(3 )

12(9 )(3 )

x x x

x x

x x

x x

And, '( ) 12[ (9 ) (3 )]V x x x

12(9 3 )

12(12 2 )

24(6 )

x x

x

x

Now, '( ) 0 9 or 3v x x x

If x = 9, then the length and the breadth will become 0.

9

3

x

x

Now, ''(3) 24(6 3) 72 0V

By second derivative test, x = 3 is the point of maxima of V.

Hence, if we remove a square of side 3 cm from each corner of the square tin and make a box

from the remaining sheet, then the volume of the box obtained is the largest possible.

Question 18:

A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off

square from each corner and folding up the flaps. What should be the side of the square to be

cut off so that the volume of the box is the maximum possible?

Solution 18:

Let the side of the square to be cut be x cm. Then, the height of the box is x, the length is 45 –

2x, ad the breadth is 24 – 2x

Therefore, the volume V(x) of the box is given by,

V(x) = x(45 – 2x)(24 – 2x)

= x(1080 – 90x – 48x + 4x2)

= 4x3 – 138x2 + 1080x

V’(x) = 12x2 – 276 + 1080

= 12(x2 – 23x +90)

= 12(x – 18)(x – 5)

V’’(x) = 24x – 276 = 12(2x – 23)

Now, V’(x) = 0 x = 18 and x =5

It is not possible to cut off a square of side 18 cm from each corner of the rectangular sheet,

Thus x cannot be equal to 18.

x = 5

Now, V’’(5) = 12(10 – 23) = 12(–13) = –156 < 0

By second derivative test x = 5 is the point of maxima.

Hence, the side of the square to be cut off to make the volume of the box maximum possible is

5 cm.



Question 19:

Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum

area.

Solution 19:

Let a rectangle of length l and breadth b be inscribed in the given circle of radius a.

Then, the diagonal passes through the center and is of length 2a cm.

Now, by applying the Pythagoras theorem, we have:

2 2 2

2 2 2

2 2

(2 )

4

4

a l b

b a l

b a l

Area of the rectangle, 2 24A l a l 2

2 2 2 2

2 2 2 2

2 2

2 2

1 14 ( 2 ) 4

2 4 4

4 2

4

dAa l l l a l

dl a l a l

a l

a l

2 2 2 2

2 2 2

2 2 2

( 2 )4 ( 4 ) (4 2 )

2 4

(4 )

la l l a l

d A a l

dl a l

2 2 2 2

3

2 2 2

2 3 2 2

3 3

2 2 2 22 2

(4 )( 4 ) 1(4 2 )

(4 )

12 2 2 (6 )

(4 ) (4 )

a l l a l

a l

a l l l a l

a l a l

Now, 0dA

dl gives 2 24 2 2a l l a

2 2 24 2 2 2b a a a a

Now, when 2l a ,



2 2 2 3

2 3 3

2( 2 )( 6 2 ) 8 24 0

2 2 2 2

d A a a a a

dl a a

By the second derivative test, when 2l a , then the area of the rectangle is the maximum.

Since 2l b a , the rectangle is a square.

Hence, it has been proved that of all the rectangles inscribed in the given fixed circle, the square

has the maximum area.

Question 20:

Show that the right circular cylinder of given surface and maximum volume is such that is

heights is equal to the diameter of the base.

Solution 20:

Let r and h be the radius and height of the cylinder respectively.

Then, the surface area (S) of the cylinder is given by, 22 2S r rh

22

2

S rh

r

1

2

Sr

r

Let V be the volume of the cylinder. Then,

2 2 31 1

2 2 2

S Sr SV r h r r r r

r r

Then, 23

2

dV Sr

dr ,

2

26

d Vr

dr

Now, 2 20 3

2 6

dV S Sr r

dr

When 2

6

Sr

, then

2

26 0

6

d V S

dr

.

By Second derivative test, the volume is the maximum when 2

6

Sr

.

Now, when 2

6

Sr

, then

26 13 2

2

rh r r r r

r

Hence, the volume is the maximum when the height is twice the radius i.e., when the height is

equal to the diameter.

Question 21:

Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimeters,

find the dimensions of the can which has the minimum surface area?

Solution 21:



Let r and h be the radius and height of the cylinder respectively.

Then, volume (V) of the cylinder is given by,

V = r2h = 100 (given)

2

100h

r

Surface area (S) of the cylinder is given by,

2 2 2002 2 2S r rh r

r

2

2004

dSr

dr r ,

2

2 3

4004

d S

dr r

2

2000 4

dSr

dr r

3 200 50

4r

1

350r

Now, it is observed that when

1

350r

, 2

20

d S

dr .

By second derivative test, the surface area is the minimum when the radius of the cylinder is 1

350

cm.

When r =

1

350

, h =

1

3502

cm.

Hence the required dimensions of the can which has them minimum surface area is given

1

350

and height =

1

3502

cm.

Question 22:

A Wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square

and the other into a circle. What should be the length of the two pieces so that the combined area

of the circle is minimum?

Solution 22:

Let a piece of length l be cut from the given wire to make a square.

Then, the other piece of wire to be made into a circle is of length (28 – l)m.

Now, side of square= 4

l



Let r be the radius of the circle. Then, 1

2 28 (28 )2

r l r l

.

The combined areas of the square and the circle (A) is given by,

A = (side of the square) 2 2r

22

22

2

2

1(28 )

16 2

1(28 )

16 4

2 2 1(28 )( 1) (28 )

16 4 8 2

10

8 2

ll

ll

dA l ll l

dl

d A l

dl

Now, 1

0 (28 ) 08 2

dA ll

dl

4(28 )0

8

( 4) 112 0

112

4

l l

l

l

Thus, when 112

4l

,

2

20

d A

dl

By second derivative test, the area (A) is the minimum when 112

4l

.

Hence, the combined area is the minimum when the length of the wire in making the square is

112

4 cm while the length of the wire in making the circle is

112 2828

4 4

cm.

Question 23:

Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8

27 of

the volume of the sphere.

Solution 23:

Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R.



Let V be the volume of the cone.

Then, 21

3V r h

Height of the cone is given by,

2 2h R AB R R r [ABC is a right triangle]

2 2 21( )

3V r R R r

2 2 2 21 1

3 3r R r R r

2 2 2

2 2

2 2 1 ( 2 )

3 3 3 2

dV rrR r R r r

dr R r

32 2

2 2

2 2 3

2 2

2 3

2 2

2 2 1

3 3 3

2 2 ( )

3 3

2 2 3

3 3

rrR r R r

R r

r R r rrR

R r

rR rrR

R r

2 2 2 2 2 3

2 2 2

2 2 2

( 2 )3 (2 9 ) (2 3 )

2 2

3 9( )

rR r R r rR r

d V R R r

dr R r

2 2 2 2 2 2 4

3

2 2 2

2 9( )(2 9 ) 2 3

327( )

R r R r r R rrR

R r

Now, 3 2

2 2

2 3 20

3 3

dV r RrR

dr R r

3 22 2 2 2

2 2

2 2 2 2 2 2

4 2 2 4 4 2 2

4 2 2

2 2

3 22 2 3 2

4 ( ) (3 2 )

4 4 9 4 12

9 8

8

9

r RR R R r r R

R r

R R r r R

R R r r R r R

r R r

r R

when 2 28

9r R , then

2

20

d V

dr

By second derivative test, the volume of the cone is the maximum when 2 28

9r R .

When 2 28

9r R ,

2 2 28 1 4

9 9 3 3

Rh R R R R R R R

Therefore,



21 8 4

3 9 3R R

38 4

27 3R

8

27 (Volume of the sphere)

Hence, the volume of the largest cone that can be inscribed in the sphere is 8

27 the volume of

the sphere.

Question 24:

Show that he right circular cone of least curved surface and given volume has an altitude equal

to 2 time the radius of the base.

Solution 24:

Let r and h be the radius and the height (altitude) of the cone respectively.

Then, the volume (V) of the cone is given as:

2

2

1 3

3

VV r h h

r

The surface area (S) of the cone is given by,

S = rl (Where l is the slant height)

2 2

2 2 2 22

2 4 2

2 6 2

9 9

19

r r h

V r r Vr r

r r

r Vr

2 52 6 2

2 6 2

2

69

2 9

rr r V

dS r V

dr r

2 6 2 6 2

2 2 6 2

2 6 2

2 2 6 2

3 9

9

2 9

9

r r V

r r V

r V

r r V

Now, 2

2 6 2 6

2

90 2 9

2

dS Vr V r

dr

Thus, it can be easily verified that when 2 2

6

2 2

9, 0

2

V d Sr

dr .

By second derivative test, the surface area of the cone is the least when 2

6

2

9

2

Vr



When, 2

6

2

9

2

Vr

,

12 6 32

2 2 2

3 3 2 3 22

9 3

V V r rh r

r r r

Hence, for a given volume, the right circular cone of the least curved surface has an altitude

equal to 2 times the radius of the base.

Question 25:

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height

is tan-1 2 .

Solution 25:

Let be the semi-vertical angle of the cone.

It is clear that 0,2

.

Let r, h and l be the radius, height, and the slant height of the cone respectively.

The slant height of the cone is given as constant.

Now r = l sin and h = l cos

The volume (V) of the cone is given by,

2

2 2

3 2

1

3

1( sin )( cos )

3

1sin cos

3

V r h

l l

l

32[sin ( sin ) cos (2sin cos )]

3

dV l

d

33 2[ sin 2sin cos ]

3

l

2 32 3 2

2[ 3sin cos 2cos 4sin cos ]

3

d V l

d

32 2[2cos 7sin cos ]

3

l

Now, 0dV

d



3 2

2

1

sin 2sin cos

tan 2

tan 2

tan 2

Now, when 1tan 2 , then 2tan 2 or 2 2sin 2cos

Then, we have: 2 3

3 3 3 3

2[2cos 14cos ] 4 cos 0

3

d V ll

d

for 0,

2

By second derivative test, the volume (V) is the maximum when 1tan 2 .

Hence, for a given slant height, the semi-vertical angle of the cone of the maximum volume is 1tan 2

Question 26:

The point on the curve x2 = 2y which is nearest to the point (0, 5) is

(A) (2 2,4) (B) (2 2,0) (C) (0, 0) (D) (2, 2)

Solution 26:

The given curve is x2 = 2y.

For each value of x, the position of the point will be 2

,2

xx

.

The distance d(x) between the points 2

,2

xx

and (0, 5) is given by,

22 4 4

2 2 2 2( ) ( 0) 5 25 5 4 252 4 4

x x xd x x x x x

3 3

4 4 22

( 8 ) ( 8 )'( )

16 1002 4 25

4

x x x xd x

x x xx

Now, 3'( ) 0 8 0d x x x

2( 8) 0

0, 2 2

x x

x

And

32 2 2 3

4 2

2 2

4 3216 100(3 8) ( 8 )

2 16 100''( )

( 16 100)

x xx x x x x

x xd x

x x

4 2 2 3 3

34 2 2

( 16 100)(3 8) 2( 8 )( 8 )

( 16 100)

x x x x x x x

x x



4 2 2 3 2

34 2 2

( 16 100)(3 8) 2( 8 )

( 16 100)

x x x x x

x x

When, x = 0 then 3

36( 8)''( ) 0

6d x

When, 2 2, ( ) 0x d x

By second derivative test, d(x) is the minimum at 2 2x .

When, 2(2 2)

2 2, 42

x y .

Hence, the point on the curve x2 = 2y which is nearest to the point (0, 5) is ( 2 2, 4) .


Question 27:

For all real values of x, the minimum value of 2

2

1

1

x x

x x

is

(A) 0 (B) 1 (C) 3 (D) 1

3

Solution 27:

Let 2

2

1( )

1

x xf x

x x

2 2

2 2

(1 )( 1 2 ) (1 )(1 2 )'( )

(1 )

x x x x x xf x

x x

2 2

2 2 2 2

2 2 2( 1)

(1 ) (1 )

x x

x x x x

2'( ) 0 1 1f x x x

Now, 2 2 2

2 4

2[(1 )(2 ) ( 1)(2)(1 )(1 2 )]''( )

(1 )

x x x x x x xf x

x x

2 2 2

2 4

2 3 2 3

2 3

3

2 3

4(1 )[(1 ) ( 1)(1 2 )]

(1 )

[ 2 1 2 ]4

(1 )

(1 3 )4

(1 )

x x x x x x x

x x

x x x x x x

x x

x x

x x

And, 3 3

4(1 3 1) 4(3) 4''(1) 0

(1 1 1) (3) 9f

And, 3

4(1 3 1)''( 1) 4( 1) 4 0

(1 1 1)f

By second derivative test f is the minimum at x = 1 and the minimum value is given by



1 1 1 1(1)

1 1 1 3f


Question 28:

The maximum value of 1

3[ ( 1) 1]x x , 0 1x is

(A)

1

31

3

(B) 1

2 (C) 1 (D) 0

Solution 28:

Let 1

3( ) [ ( 1) 1]f x x x .

2

3

2 1'( )

3[ ( 1) 1]

xf x

x x

Now, 1

( ) 02

f x x

But, x = -1/2 is not part of the interval [0,1]

Then, we evaluate the value of f at the end points of the interval [0, 1] {i.e., at x = 0 and x =1}. 1

3

1

3

(0) [0(0 1) 1] 1

(0) [1(1 1) 1] 1

f

f

Hence, we can conclude that the maximum value of f in the interval [0, 1] is 1.

The correct answer is C.

Miscellaneous Exercise

Question 1:

Using differentials, find the approximate value of each of the following.

(a)

1

417

81

(b) 1

533

Solution 1:

(a) Consider 1

4y x . Let 16

81x and

1

81x .

Then, 1 1

4 4( )y x x x



1 1

4 4

1

4

1

4

17 16

81 81

17 2

81 3

17 2

81 3y


3

4

1( )

4( )

dydy x x

dxx

1

4as y x

3

4

1 1 27 1 1 10.010

81 4 8 81 32 3 9616

481

Hence, the approximate value of

1

417

81

is

20.010 0.667 0.010

3

= 0.677.

(b) Consider 1

5y x . Let x =32 and 1x .

Then,

1 1 1 1 1

5 5 5 5 51

( ) (33) (32) (33)2

y x x x

1

51

(33)2

y


6

5

1( )

5( )

dydy x x

dxx

1

5as y x

6

1 11 0.003

3205 2


533 is 1

( 0.003) 0.5 0.003 0.4972 .

Question 2:

Show that the function given by log

( )x

f xx

has maximum at x = e.

Solution 2:



The given function is log

( )x

f xx

2 2

1log

1 log'( )

x xxx

f xx x

Now, f’(x) = 0

1 – log x = 0

log x = 1

log x = log e

x = e

Now,

2

4

1(1 log )(2 )

''( )

x x xx

f xx

4

3

2 (1 log )

3 2log

x x x

x

x

x

Now, 3 3 3

3 2log 3 2 1''( ) 0

ef e

e e e

Therefore, by second derivative test f is the maximum at x = e.

Question 3:

The two equal sides of an isosceles triangle with fixed base are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?

Solution 3:

Let ∆ABC be isosceles where BC is the base of fixed length b.

Let the length of the two equal sides of ∆ABC be a.

AD BC.

Draw

Now, in ∆ADC, by applying the Pythagoras theorem, we have:

22

4

bAD a

b



Area of triangle 2

21( )

2 4

bA b a

The rate of change of the area with respect to time (t) is given by,

2 2 22

1 2

2 42

4

dA a da ab dab

dt dt dtb a ba

It is given that the two equal sides of the triangle are decreasing at the rate of 3 cm per second.

2 2

3 cm/S

3

4

da

dt

da ab

dt a b

Then, when a = b,we have: 2 2

2 2 2

3 33

4 3

dA ab bb

dt b b b

Hence, if two equal sides are equal to the base, then the area of the triangle is decreasing at the

rate of 23 cm /s.b

Question 4:

Find the equation of the normal to curve 2 4y x at the point (1.2).

Solution 4:


Differentiating with respect to x, we have:

(1,2)

2 4

4 2

21

2

dyy

dx

dy

dx dy y

dy

dx

Now, the slope of the normal at point (1, 2) is

(1,2)

1 11

1dy

dx

.

Equation of the normal at (1, 2) is y – 2 = –1(x – 1).

y – 2 = –x + 1

x + y – 3 = 0

Question 5:

Show that the normal at any point θ to the curve

cos sin , sin cosx a a y a a is at a constant distance from the origin.



Solution 5:

We have x = a cos θ + aθ sin θ

sin sin cos cos

sin cos

cos cos sin

. tancos

dxa a a a

d

y a a

dya a a sin a

d

dy dy d a sin

dx d dx a

Slope of the normal at any point 1

.tan

The equation of the normal at a given point (x,y)is given by,

2 2

2 2

1sin cos ( cos sin )

tan

sin sin sin cos cos cos sin cos

cos sin (sin cos ) 0

cos sin 0

y a a x a a

y a a x a

x y a

z y a

Now, the perpendicular distance of the normal from the origin is

2 2

| | | || |

1cos sin

a aa

, which is independent of θ.

Hence, the perpendicular distance of the normal from the origin is constant.

Question 6:

Find the intervals in which the function given f by

4sin 2 cos( )

2 cos

x x x xf x

x

Is (i) increasing (ii) decreasing

Solution 6:

2

2 2 2

4sin 2 cos( )

2 cos

(2 cos )(4 cos 2 cos sin ) (4sin 2 cos )( sin )'( )

2 cos

(2 cos )(3cos 2 sin ) sin (4sin 2 cos )

(2 cos )

6cos 4 2 sin 3cos 2cos sin cos 4sin 2sin 2 s

x x x xf x

x

x x x x x x x x x x xf x

x

x x x x x x x x x

x

x x x x x x x x x x

2

2

2 2

in sin cos

(2 cos )

4cos cos cos (4 cos )

(2 cos ) (2 cos )

x x x x

x

x x x x

x x

= x

Now, '( ) 0f x

cos x = 0, cos x = 4



But, cos x 4

cos x = 0

3,

2 2x

Now, 2

x

and 3

2x

divides (0, 2π) into three disjoint intervals i.e.,

0,2

, 3

,2 2

, and 3

, 22

In intervals 0,2

and 3

, 22

, '( ) 0f x .

Thus, f(x) is increasing for 0 < x < 2

x and

3

2

< x < 2 .

In the interval, 3

,2 2

, '( ) 0f x .

Thus, f(x) is decreasing for 2

< x <

3

2

.

Question 7:

Find the intervals in which the function f given by f(x) = x3 + 3

1

x, x 0 is

(i) increasing (ii) decreasing

Solution 7:

f(x) = x3 + 3

1

x

62

4 4

3 3 3'( ) 3

xf x x

x x

Then, f’(x) = 0 3x6 – 3 = 0 x6 = x = 1 .

Now, the points x = 1 and x = –1 divide the real line into three disjoint intervals

i.e., ( , 1) , ( 1, 1) and (1, ) .

In intervals ( , 1) and (1, ) i.e., when x < –1 and x > 1, '( ) 0f x .

Thus, when x < –1 and x > 1, is increasing.

In interval ( 1, 1) i.e., when –1 < x < 1, '( ) 0f x .

Thus, when –1 < x < 1, f is decreasing.

Question 8:

Find the maximum area of an isosceles triangle inscribed in the ellipse 2 2

2 21

x y

a b with its

vertex at one end of the major axis.

Solution 8:

f



The given ellipse is 2 2

2 21

x y

a b

Let the major axis be along the x-axis.

Let ABC, be the triangle inscribed in the ellipse where vertex C is at (a, 0).

Since the ellipse is symmetrical with respect to the x-axis and y-axis, we can assume the

coordinates of A to be (–x1, y1).

Now, we have 2 2

1 1

by a x

a .

Coordinates of A are 2 2

1 1,b

x a xa

and the coordinates of B are 2 2

1 1,b

x a xa

.

As the point (–x1, y1), lies on the ellipse, the area of triangle ABC (A) is given by,

2 2 2 2 2 2

1 1 1 1 1

2 2 2 2

1 1 1

1 2( ) ( )

2

b b bA a a x x a x x a x

a a a

bA ba a x x a x

a

…… (1)

22 2 1

12 2 2 2 2

1 1 1

2 2 2

1 1 12 2

1

2 2 2

1 1

2 2

1

22

2

[ ( ) ]2

( 2 )

2

bxdA xb ba x

dx aa x a a x

bx a a x x

a x

b x x a

a x

Now, 1

0dA

dx



2 2

1 1

2 2

1

2

1

2 0

4( 2)( )

2( 2)

9

4

3

4

,2

x x a a

a a ax

a a

a a

ax a

But x1 cannot be equal to a. 2

2

1 1

33

2 4 2 2

a b a ba bx y a

a a

Now,

2 2 2 2 11 1 1 1 2 22

1

2 2 2

1 1

2( 4 ) ( 2 )

2

xa x x a x x a a

a xd A b

dx a a x

2 2 2 2

1 1 1 1 1

2

2 2 31

3 2 3

3

2 2 21

( )( 4 ) ( 2 )

( )

2 3 )

( )

b a x x a x x x a a

aa x

b x a x a

aa x

Also, when 1

2

ax , then

3 3 33 3 3

2

3 322 21 2 2

3

32 2

32 3

8 2 4 2

3 3

4 4

9

4 0

3

4

a a aa a a

d A b b

dx a aa a

ab

aa

Thus, the area is the maximum when 1

2

ax .

Maximum area of the triangle is given by,



2 22 2

4 2 4

3 3

2 2 2

3 3 3 3

2 4 4

a a b aA b a a

a

a b aab

a

ab abab

Question 9:

A tank with rectangular base and rectangular sides, open at the top is to constructed so that its

depth is 2 m and volume is 8 m3. If building of tank costs Rs. 70 per sq. meters for the base and

Rs. 45 per sq. meters for sides. What is the cost of least expensive tank?

Solution 9:

Let l, b and h represent the length, breadth, and height of the tank respectively.

Then, we have height (h) = 2 m

Volume of the tank = 8 m3

Volume of the tank = lbh

8 = l × b × 2

44lb b

l

Now, area of the base = lb = 4

Area, of the 4 walls (A) = 2h(l + b)

2

44

44

A ll

dAl

dl l

Now, 0dA

dl

2

2

40

4

2

ll

l

l

However, the length cannot be negative.

Therefore, we have l = 4.

4 42

2b

l

Now, 2

2 3

32d A

dl l

When, 2l



2

2

324 0

8

d A

dl

Thus, by second derivative test, the area is the minimum when l = 2.

We have l = b = h = 2.

Cost of building the base = Rs. 70 × (lb) = Rs. 70(4) = Rs. 280

Cost of building the walls = Rs. 2h (l + h) × 45 = Rs. 90(2)(2 + 2) = Rs. 8(90) = Rs. 720

Required total cost = Rs. (280 + 720) = Rs. 1000

Hence, the total cost of the tank will be Rs. 1000.

Question 10:

The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the

sum of their area is least when the side of square is double the radius of the circle.

Solution 10:

Let r be the radius of the circle and a be the side of the square.

Then, we have:

2πr + 4a = k (where k is constant)

2

4

k ra

The sum of the areas of the circle and the square (A) is given by, 2

2 2 2 ( 2 )

16

2( 2 )(2 ) ( 2 )2 2

16 4

k rA r a r

dA k r k rr r

dr

Now, 0dA

dr

( 2 )2

4

8 2

(8 2 )

8 2 2(4 )

k rr

r k r

r k

k kr

Now, 2 2

22 0

2

d A

dr

Where 2

2, 0

2(4 )

k d Ar

dr

The sum of the areas is least when 2(4 )

kr

where

22(4 ) 8 2 2

, 22(4 ) 4 2(4 ) 4 4

kk

k k k k kr a r



Hence, it has been proved that the sum of their areas is least when the side of the square is double

the radius of the circle.

Question 11:

A window is in the form of rectangle surmounted by a semicircular opening. The total perimeter

of the window is 10 m. Find the dimensions of the window to admit maximum light through the

whole opening.

Solution 11:

Let x and y be the length and breadth of the rectangular window.

Radius of the semicircular opening = 2

x

It is given that the perimeter of the window is 10 m.

2 102

1 2 102

2 10 12

15

2 4

xx y

x y

y x

y x

Area of the window (A) is given by, 2

2

2 2

2

2

2 2

15

2 4 8

15

2 4 8

15 2

2 4 4

1 12 4 4

xA xy

x x x

x x x

dAx x

dx

d A

dx

Now, 0dA

dx



5 1 02 4

5 04

1 54

x x

x x

x

5 20

41

4

x

Thus, when 20

4x

then

2

20

d A

dx .

Therefore, by second derivative test, the area is the maximum when length 20

4x

m.

Now,

20 2 5(2 ) 105 5 m

4 4 4 4y

Hence, the required dimensions of the window to admit maximum light is given by

length = 20

m4

and breadth = 10

m4

.

Question 12:

A point of the hypotenuse of a triangle is at distance and from the sides of the triangle. Show

that the minimum length of the hypotenuse is

32 2 23 3a b

Solution 12:

Let ∆ABC be right-angles at B. Let AB = x and BC = y.

Let P be a point on the hypotenuse of the triangle such that P is at a distance of a and b from the

sides AB and BC respectively.

Let ∠c = θ

We have, 2 2AC x y

a b



Now,

PC = b cosec θ

And, AP = a sec θ

AC = AP + PC

AC = b cosec θ + a sec θ … (1)

( )cos cot sec tan

( )0

d ACb ec a

d

d AC

d

3 3

1 1

3 3

1

3

sec tan cosec cot

sin cos

cos cos sin sin

sin cos

( ) sin ( ) cos

tan

a b

a b

a b

a b

b

a

1

3

2 2

3 3

( )sin

b

a b

and

1

3

2 2

3 3

( )cos

a

a b

…(2)

It can be clearly shown that 2

2

( )0

d AC

d when

1

3

tanb

a

.

Therefore, by second derivative test, the length of the hypotenuse is the maximum when 1

3

tanb

a

.

Now, when

1

3

tanb

a

, we have:

1

3

2 2 2 2

3 3 3 3

1 1

3 3

tanb

a

b a b a a bAC

b a

[Using (1) and (2)]

2 2 2 2

3 3 3 3

32 2 23 3

a a b b a

a b



Hence, the maximum length of the hypotenuses is =

32 2 23 3a b

.

Question 13:

Find the points at which the function f given by f(x) = (x – 2)4(x + 1)3 has

(i) local maxima (ii) local minima (iii) point of inflexion

Solution 13:

The given function is f(x) = (x – 2)4(x + 1)3

3 3 2 4' 4 – 2) 1( ) ( ( ) 3( )1 )2(f x x x xx

3 2

3 2

– 2) 1 1 2

– 2) 1 2

( ( ) [4( ) 3( )]

( ( ) (7 )

x x x x

x x x

Now, f’(x) = 0 x = -1 and 2

7x or x = 2

Now, for values of x close to 2

7 and to the left of

2

7, f(x) > 0. Also, for values of x close to

2

7

and to the right of 2

7, f’(x) > 0.

Thus, 2

7x is the point of local minima.

Now, as the value of x varies through -1 f’(x) does not changes its sign.

Thus, x = -1 is the point of inflexion.

Question 14:

Find the absolute maximum and minimum values of the function f given by

f(x) = cos2 x + sin x, [0, ]x

Solution 14:

f(x) = cos2 x + sin x

'( ) 2cos ( sin ) cosf x x x x

= –2sin x cos x + cos x

Now, f'(x) = 0

2sin cos cos cos (2sin 1) 0

1sin or cos 0

2

or as [0, ]6 2

x x x x x

x x

x x

Now, evaluating the value of f at critical points x = 2

and x =

6

and at the end points of the

interval [0, ] (i.e., at x = 0 and x = n), we have:



2

2

2

22

2

3 1 5cos sin

6 6 6 2 2 4

0 cos 0 sin 0 1 0 1

cos sin 1 0 1

cos sin 0 1 12 2 2

f

f

f

f

Hence, the absolute maximum value of f is 5

4 occurring at x =

6

and the absolute minimum

value of f is 1 occurring at x = 0, x = 2

, and π.

Question 15:

Show that the altitude of the right circular cone of maximum volume that can be inscribed in a

sphere of radius r is 4

3

r.

Solution 15:

A sphere of fixed radius (r) is given.

Let R and h be the radius and the height of the cone respectively.

The volume (V) of the cone is given by,

21

3V R h

Now, from the right triangle BCD, we have:

2 2

2 2

BC r R

h r r R



2 2 2 2 2 2 2

22 2

2 2

32 2

2 2

2 2 3

2 2

2 3

2 2

1 1 1( )

3 3 3

2 2 ( 2 )

3 3 3 2

2 2

3 3 3

2 2 ( )

3 3

2 2 3

3 3

V R r r R R r R r R

dV R RRr R r R

dR r R

RRr R r R

r R

Rr r R RRr

r R

Rr RrRr

r R

Now, 2

20

d V

dR

2 2

2 2

2 2 2 2

2 2 2 2 2 2

4 2 2 4 4 2 2

4 2 2

2 2

22

2 3 2

3 3

2 3 2

4 ( ) (3 2 )

14 4 9 4 12

9 8 0

9 8

8

9

rR R Rr

r R

r r R R r

r r R R r

r r R R r R r

R R r

R r

rR

Now,

2 2 2 2 3 3

22 2 2

2 2 2

13 (2 9 ) (2 3 )( 6 )

2 2

3 9( )

r R r R R R Rd V r r R

dR r R

2 2 2 2 3 3

22 2

2 2

13 (2 9 ) (2 3 )(3 )

2 2

3 9( )

r R r R R R Rr r R

r R

Now, when 2

2 8

9

rR , it can be shown that

2

20

d V

dR .

The volume is the maximum when 2

2 8

9

rR .

When 2

2 8

9

rR , height of the cone =

2 22 8 4

9 9 3 3

R r r rr r r r .

Hence, it can be seen that the altitude of the right circular cone of maximum volume that can be

inscribed in a sphere of radius r is 4

3

r.



Question 17:

Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of

radius R is 2

3

R, also find the maximum volume.

Solution 17:

A sphere of fixed radius (R) is given.

Let r and h be the radius and the height of the cylinder respectively.

Form the given figure, we have 2 22h R r

The volume (V) of the cylinder is given by,

2 2 2 2

22 2

2 2

32 2

2 2

2 2 3

2 2

2 3

2 2

2

2 ( 2 )4

2

24

4 ( ) 2

4 6

V r h r R r

dV r rr R r

dr R r

rr R r

R r

r R r r

R r

rR r

R r

Now, 2 30 4 6 0

dVrR r

dr

22 2

3

Rr

Now,

2 2 2 2 2 3

2 2 2

2 2 2

( 2 )(4 18 ) (4 6 )

2

( )

rR r R r rR r

d V R r

dr R r

2 2 2 2 2 3

32 2 2

4 2 2 4 2 2

3

2 2 2

( )(4 18 ) (4 6 )

( )

4 22 12 4

( )

R r R r r rR r

R r

R r R r r R

R r

Now, it can be observed that at 2

2 2

3

Rr ,

2

20

d V

dr .



The volume is the maximum when 2

2 2

3

Rr

When 2

2 2

3

Rr the height of the cylinder is

2 22 2 2

2 23 3 3

R R RR .

Hence, the volume of the cylinder is the maximum when the height of the cylinder is 2

3

R.

Question 18:

Show that height of the cylinder of greatest volume which can be inscribed in a right circular

cone of height h and semi vertical angle a is one-third that of the cone and the greatest volume

of cylinder is 2 24tan

27h a .

Solution 18:

The given right circular cone of fixed height (h) and semi-vertical angle (a) can be drawn as:

Here, a cylinder of radius R and height H is inscribed in the cone.

Then, ∆GAO = a, OG = r, OA = h, OE = R, and CE = H.

We have,

r = h tan a

Now, since ∆AOG is similar to ∆CEG, we have:

AO CE

OG EG

h H

r r R

[EG = OG – OE]

1( ) ( tan ) ( tan )

tan tan

h hH r R h a R h a R

r h a a

Now, the volume (V) of the cylinder is given by, 2 3

2 2( tan )tan tan

R RV R H h a R R h

a a

232

tan

dV RRh

dR a

Now, 0dV

dR



232

tan

2 tan 3

2tan

3

RRh

a

h a R

hR a

Now, 2

2

62

tan

d V RRh

dR a

And, for 2

tan3

hR a we have:

2

2

6 22 tan 2 4 2 0

tan 3

d V hh a h h h

dR a

By second derivative test, the volume of the cylinder is the greatest when 2

tan3

hR a .

When 2

tan3

hR a ,

1 2 1 tantan tan

tan 3 tan 3 3

h h a hH h a a

a a

Thus, the height of the cylinder is one-third the height of the cone when the volume of the

cylinder is the greatest.

Now, the maximum volume of the cylinder can be obtained as: 22 2

2 3 22 4 4tan tan tan

3 3 9 3 27

h h h ha a h a

Hence, the given result is proved.

Question 19:

A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic mere per

hour. Then the depth of the wheat is increasing at the rate of

(A) 1 m/h (B) 0.1 m/h (C) 1.1 m/h (D) 0.5 m/h

Solution 19:

Let r be the radius of the cylinder.

Then, volume (V) of the cylinder is given by,

V = π(radius)2 × height

= π(10)2h (radius = 10 m)

= 100 πh

Differentiating with respect to time t, we have:

100dt

dV

dt

dh

The tank is being filled with wheat at the rate of 314 cubic meters per hour.

3314 m /hdV

dt

Thus, we have:



314 100

314 3141

100(3.14) 314

dh

dt

dh

dt

Hence, the depth of wheat is increasing at the rate of 1 m/h.


Question 20:

The slope of the tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point (2, –1) is

(A) 22

7 (B)

6

7 (C)

7

6 (D)

6

7

Solution 20:

The given curve is x = t2 + 3t – 8, and y = 2t2 – 2t – 5

2 3dx

tdt

and 4 2dy

tdt

4 2

2 3

dy dy dt t

dx dt dx t

The given points is (2, –1).

At x = 2, we have:

t2 + 3t – 8 = 2

t2 + 3t – 10 = 0

(t – 2)(t + 5) = 0

t = 2 or t = –5

At y = –1, we have

2t2 – 2t – 5 = –1

2t2 – 2t – 4 = 0

2(t2 – t – 2) = 0

(t – 2)(t + 1) = 0

t = 2 or t = –1

The common value of t is 2.

Hence, the slope of the tangent to the given curve at point (2, –1) is

2

4(2) 2 8 2 6

2(2) 3 4 3 7t

dy

dx


Question 21:

The line y = mx + 1 is tangent to the given curve y2 = 4x if the value on m is

(A) 1 (B) 2 (C) 3 (D) 1

2

Solution 21:



The equation of the tangent to the given curve is y = mx + 1

Now, substituting y = mx + 1 in y2 = 4x, we get:

(mx + 1)2 = 4x

m2x2 + 1 + 2mx – 4x = 0

m2x2 + x(2m – 4) + 1 = 0 ….(i)

Since a tangent touches the curve at one point, the toots of equation (i) must be equal.

Therefore, we have:

Discriminant = 0

(2m – 4)2 – 4(m2)(1) = 0

4m2 + 16 – 16m – 4m2 = 0

16 – 16m = 0

m = 1

Hence, the required value of m is 1.


Question 22:

The normal at the point (1, 1) on the curve 2y + x2 = 3 is

(A) x + y = 0 (B) x – y = 0 (C) x + y + 1 = 0 (D) x – y = 1

Solution 22:

The equation of the given curve is 2y + x2 = 3


(1,1)

22 0

1

dyx

dx

dyx

dx

dy

dx

The slope of the normal to the given curve at point (1, 1) is

(1,1)

11

dy

dx

Hence, the equation of the normal to the given curve at (1, 1) is given as:

y – 1 = 1(x – 1)

y – 1 = x – 1

x – y = 0


Question 23:

The normal to the curve x2 = 4y passing (1, 2) is

A) x + y = 3 (B) x – y = 3 (C) x + y = 1 (D) x – y = 1



Solution 23:

The equation of the given curve is x2 = 4y.


2 4

2

dyx

dx

dy x

dx

The slope of the normal to the given curve at point (h, k) is given by,

( , )

1 2

h k

dy h

dx

Equation of the normal at point (h, k) is given as:

2( )y k x h

h

Now, it is given that the normal passes through the point (1, 2).

Therefore, we have:

2 22 (1 ) or 2 (1 )k h k h

h h

…(i)

Since (h, k) lies on the curves x2 = 4y, we have h2 = 4k 2

4

hk

From equation (i), we have: 2

3

3

2

22 (1 )

4

2 2 2 24

8

2

14

hh

h

hh h

h

h

hk k

Hence, the equation of the normal is given as:

21 ( 2)

2

1 ( 2)

3

y x

y x

x y


Question 24:

The points on the curve 9y2 = x3, where the normal to the curve makes equal intercepts with the

axes are



(A) 8

4,3

(B)

84,

3

(C) 3

4,8

(D)

84,

3

Solution 24:

The equation of the given curve is 9y2 = x2


29(2 ) 3dy

y xdx

2

6

dy x

dx y

The slope of the normal to the given curve at point (x1, y1) is

1 1

1

2

1

( , )

61

x y

y

dy x

dx

The equation of the normal to the curve at (x1, y1) is

11 12

1

2 2

1 1 1 1 1 1

2 2

1 1 1 1 1 1 1

2

1

2 2

1 1 1 1 1 1 1

1 1 1 1

1

6( )

6 6

6 6

61

6 6

1(6 ) (6 )

6

yy y x x

x

x y x y xy x y

x y x y x y x y

xy x y

x y x y x y x y

x y

x x y x

x

It is given that the normal makes equal intercepts with the axes.

Therefore, we have:

1 1 1 1

1

1 1

1

2

1 1

(6 ) (6 )

6

6

6 ....( )

x x y x

x

x y

x

x y i

Also, the point (x1, y1) lies on the curve, so we have

9y12 = x1

3 …(ii)

From (i) and (ii), we have: 2

2 43 31 1

1 1 19 46 4

x xx x x

From (iii), we have:



2 3

1

2

1

1

9 (4) 64

64

9

8

3

y

y

y

Hence, the required points are 8

4,3

.



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Class XII Maths Chapter-06 - Application of Derivatives...Class XII – NCERT – Maths Chapter-06 - Application of Derivatives 6. Application of Derivatives Exercise 6.1. Question